-0.000 000 000 000 176 557 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 82(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 82(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 82| = 0.000 000 000 000 176 557 82


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 82.

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 82 × 2 = 0 + 0.000 000 000 000 353 115 64;
  • 2) 0.000 000 000 000 353 115 64 × 2 = 0 + 0.000 000 000 000 706 231 28;
  • 3) 0.000 000 000 000 706 231 28 × 2 = 0 + 0.000 000 000 001 412 462 56;
  • 4) 0.000 000 000 001 412 462 56 × 2 = 0 + 0.000 000 000 002 824 925 12;
  • 5) 0.000 000 000 002 824 925 12 × 2 = 0 + 0.000 000 000 005 649 850 24;
  • 6) 0.000 000 000 005 649 850 24 × 2 = 0 + 0.000 000 000 011 299 700 48;
  • 7) 0.000 000 000 011 299 700 48 × 2 = 0 + 0.000 000 000 022 599 400 96;
  • 8) 0.000 000 000 022 599 400 96 × 2 = 0 + 0.000 000 000 045 198 801 92;
  • 9) 0.000 000 000 045 198 801 92 × 2 = 0 + 0.000 000 000 090 397 603 84;
  • 10) 0.000 000 000 090 397 603 84 × 2 = 0 + 0.000 000 000 180 795 207 68;
  • 11) 0.000 000 000 180 795 207 68 × 2 = 0 + 0.000 000 000 361 590 415 36;
  • 12) 0.000 000 000 361 590 415 36 × 2 = 0 + 0.000 000 000 723 180 830 72;
  • 13) 0.000 000 000 723 180 830 72 × 2 = 0 + 0.000 000 001 446 361 661 44;
  • 14) 0.000 000 001 446 361 661 44 × 2 = 0 + 0.000 000 002 892 723 322 88;
  • 15) 0.000 000 002 892 723 322 88 × 2 = 0 + 0.000 000 005 785 446 645 76;
  • 16) 0.000 000 005 785 446 645 76 × 2 = 0 + 0.000 000 011 570 893 291 52;
  • 17) 0.000 000 011 570 893 291 52 × 2 = 0 + 0.000 000 023 141 786 583 04;
  • 18) 0.000 000 023 141 786 583 04 × 2 = 0 + 0.000 000 046 283 573 166 08;
  • 19) 0.000 000 046 283 573 166 08 × 2 = 0 + 0.000 000 092 567 146 332 16;
  • 20) 0.000 000 092 567 146 332 16 × 2 = 0 + 0.000 000 185 134 292 664 32;
  • 21) 0.000 000 185 134 292 664 32 × 2 = 0 + 0.000 000 370 268 585 328 64;
  • 22) 0.000 000 370 268 585 328 64 × 2 = 0 + 0.000 000 740 537 170 657 28;
  • 23) 0.000 000 740 537 170 657 28 × 2 = 0 + 0.000 001 481 074 341 314 56;
  • 24) 0.000 001 481 074 341 314 56 × 2 = 0 + 0.000 002 962 148 682 629 12;
  • 25) 0.000 002 962 148 682 629 12 × 2 = 0 + 0.000 005 924 297 365 258 24;
  • 26) 0.000 005 924 297 365 258 24 × 2 = 0 + 0.000 011 848 594 730 516 48;
  • 27) 0.000 011 848 594 730 516 48 × 2 = 0 + 0.000 023 697 189 461 032 96;
  • 28) 0.000 023 697 189 461 032 96 × 2 = 0 + 0.000 047 394 378 922 065 92;
  • 29) 0.000 047 394 378 922 065 92 × 2 = 0 + 0.000 094 788 757 844 131 84;
  • 30) 0.000 094 788 757 844 131 84 × 2 = 0 + 0.000 189 577 515 688 263 68;
  • 31) 0.000 189 577 515 688 263 68 × 2 = 0 + 0.000 379 155 031 376 527 36;
  • 32) 0.000 379 155 031 376 527 36 × 2 = 0 + 0.000 758 310 062 753 054 72;
  • 33) 0.000 758 310 062 753 054 72 × 2 = 0 + 0.001 516 620 125 506 109 44;
  • 34) 0.001 516 620 125 506 109 44 × 2 = 0 + 0.003 033 240 251 012 218 88;
  • 35) 0.003 033 240 251 012 218 88 × 2 = 0 + 0.006 066 480 502 024 437 76;
  • 36) 0.006 066 480 502 024 437 76 × 2 = 0 + 0.012 132 961 004 048 875 52;
  • 37) 0.012 132 961 004 048 875 52 × 2 = 0 + 0.024 265 922 008 097 751 04;
  • 38) 0.024 265 922 008 097 751 04 × 2 = 0 + 0.048 531 844 016 195 502 08;
  • 39) 0.048 531 844 016 195 502 08 × 2 = 0 + 0.097 063 688 032 391 004 16;
  • 40) 0.097 063 688 032 391 004 16 × 2 = 0 + 0.194 127 376 064 782 008 32;
  • 41) 0.194 127 376 064 782 008 32 × 2 = 0 + 0.388 254 752 129 564 016 64;
  • 42) 0.388 254 752 129 564 016 64 × 2 = 0 + 0.776 509 504 259 128 033 28;
  • 43) 0.776 509 504 259 128 033 28 × 2 = 1 + 0.553 019 008 518 256 066 56;
  • 44) 0.553 019 008 518 256 066 56 × 2 = 1 + 0.106 038 017 036 512 133 12;
  • 45) 0.106 038 017 036 512 133 12 × 2 = 0 + 0.212 076 034 073 024 266 24;
  • 46) 0.212 076 034 073 024 266 24 × 2 = 0 + 0.424 152 068 146 048 532 48;
  • 47) 0.424 152 068 146 048 532 48 × 2 = 0 + 0.848 304 136 292 097 064 96;
  • 48) 0.848 304 136 292 097 064 96 × 2 = 1 + 0.696 608 272 584 194 129 92;
  • 49) 0.696 608 272 584 194 129 92 × 2 = 1 + 0.393 216 545 168 388 259 84;
  • 50) 0.393 216 545 168 388 259 84 × 2 = 0 + 0.786 433 090 336 776 519 68;
  • 51) 0.786 433 090 336 776 519 68 × 2 = 1 + 0.572 866 180 673 553 039 36;
  • 52) 0.572 866 180 673 553 039 36 × 2 = 1 + 0.145 732 361 347 106 078 72;
  • 53) 0.145 732 361 347 106 078 72 × 2 = 0 + 0.291 464 722 694 212 157 44;
  • 54) 0.291 464 722 694 212 157 44 × 2 = 0 + 0.582 929 445 388 424 314 88;
  • 55) 0.582 929 445 388 424 314 88 × 2 = 1 + 0.165 858 890 776 848 629 76;
  • 56) 0.165 858 890 776 848 629 76 × 2 = 0 + 0.331 717 781 553 697 259 52;
  • 57) 0.331 717 781 553 697 259 52 × 2 = 0 + 0.663 435 563 107 394 519 04;
  • 58) 0.663 435 563 107 394 519 04 × 2 = 1 + 0.326 871 126 214 789 038 08;
  • 59) 0.326 871 126 214 789 038 08 × 2 = 0 + 0.653 742 252 429 578 076 16;
  • 60) 0.653 742 252 429 578 076 16 × 2 = 1 + 0.307 484 504 859 156 152 32;
  • 61) 0.307 484 504 859 156 152 32 × 2 = 0 + 0.614 969 009 718 312 304 64;
  • 62) 0.614 969 009 718 312 304 64 × 2 = 1 + 0.229 938 019 436 624 609 28;
  • 63) 0.229 938 019 436 624 609 28 × 2 = 0 + 0.459 876 038 873 249 218 56;
  • 64) 0.459 876 038 873 249 218 56 × 2 = 0 + 0.919 752 077 746 498 437 12;
  • 65) 0.919 752 077 746 498 437 12 × 2 = 1 + 0.839 504 155 492 996 874 24;
  • 66) 0.839 504 155 492 996 874 24 × 2 = 1 + 0.679 008 310 985 993 748 48;
  • 67) 0.679 008 310 985 993 748 48 × 2 = 1 + 0.358 016 621 971 987 496 96;
  • 68) 0.358 016 621 971 987 496 96 × 2 = 0 + 0.716 033 243 943 974 993 92;
  • 69) 0.716 033 243 943 974 993 92 × 2 = 1 + 0.432 066 487 887 949 987 84;
  • 70) 0.432 066 487 887 949 987 84 × 2 = 0 + 0.864 132 975 775 899 975 68;
  • 71) 0.864 132 975 775 899 975 68 × 2 = 1 + 0.728 265 951 551 799 951 36;
  • 72) 0.728 265 951 551 799 951 36 × 2 = 1 + 0.456 531 903 103 599 902 72;
  • 73) 0.456 531 903 103 599 902 72 × 2 = 0 + 0.913 063 806 207 199 805 44;
  • 74) 0.913 063 806 207 199 805 44 × 2 = 1 + 0.826 127 612 414 399 610 88;
  • 75) 0.826 127 612 414 399 610 88 × 2 = 1 + 0.652 255 224 828 799 221 76;
  • 76) 0.652 255 224 828 799 221 76 × 2 = 1 + 0.304 510 449 657 598 443 52;
  • 77) 0.304 510 449 657 598 443 52 × 2 = 0 + 0.609 020 899 315 196 887 04;
  • 78) 0.609 020 899 315 196 887 04 × 2 = 1 + 0.218 041 798 630 393 774 08;
  • 79) 0.218 041 798 630 393 774 08 × 2 = 0 + 0.436 083 597 260 787 548 16;
  • 80) 0.436 083 597 260 787 548 16 × 2 = 0 + 0.872 167 194 521 575 096 32;
  • 81) 0.872 167 194 521 575 096 32 × 2 = 1 + 0.744 334 389 043 150 192 64;
  • 82) 0.744 334 389 043 150 192 64 × 2 = 1 + 0.488 668 778 086 300 385 28;
  • 83) 0.488 668 778 086 300 385 28 × 2 = 0 + 0.977 337 556 172 600 770 56;
  • 84) 0.977 337 556 172 600 770 56 × 2 = 1 + 0.954 675 112 345 201 541 12;
  • 85) 0.954 675 112 345 201 541 12 × 2 = 1 + 0.909 350 224 690 403 082 24;
  • 86) 0.909 350 224 690 403 082 24 × 2 = 1 + 0.818 700 449 380 806 164 48;
  • 87) 0.818 700 449 380 806 164 48 × 2 = 1 + 0.637 400 898 761 612 328 96;
  • 88) 0.637 400 898 761 612 328 96 × 2 = 1 + 0.274 801 797 523 224 657 92;
  • 89) 0.274 801 797 523 224 657 92 × 2 = 0 + 0.549 603 595 046 449 315 84;
  • 90) 0.549 603 595 046 449 315 84 × 2 = 1 + 0.099 207 190 092 898 631 68;
  • 91) 0.099 207 190 092 898 631 68 × 2 = 0 + 0.198 414 380 185 797 263 36;
  • 92) 0.198 414 380 185 797 263 36 × 2 = 0 + 0.396 828 760 371 594 526 72;
  • 93) 0.396 828 760 371 594 526 72 × 2 = 0 + 0.793 657 520 743 189 053 44;
  • 94) 0.793 657 520 743 189 053 44 × 2 = 1 + 0.587 315 041 486 378 106 88;
  • 95) 0.587 315 041 486 378 106 88 × 2 = 1 + 0.174 630 082 972 756 213 76;

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 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0100 1110 1011 0111 0100 1101 1111 0100 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0100 1110 1011 0111 0100 1101 1111 0100 011(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 82(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0100 1110 1011 0111 0100 1101 1111 0100 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0100 1110 1011 0111 0100 1101 1111 0100 011(2) × 20 =

1.1000 1101 1001 0010 1010 0111 0101 1011 1010 0110 1111 1010 0011(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 1010 0111 0101 1011 1010 0110 1111 1010 0011


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 1010 0111 0101 1011 1010 0110 1111 1010 0011 =

1000 1101 1001 0010 1010 0111 0101 1011 1010 0110 1111 1010 0011


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 1010 0111 0101 1011 1010 0110 1111 1010 0011


Decimal number -0.000 000 000 000 176 557 82 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1010 0111 0101 1011 1010 0110 1111 1010 0011

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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