-0.000 000 000 000 176 549 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 549 7(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
-0.000 000 000 000 176 549 7(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:

|-0.000 000 000 000 176 549 7| = 0.000 000 000 000 176 549 7


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: 0.000 000 000 000 176 549 7.

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) 0.000 000 000 000 176 549 7 × 2 = 0 + 0.000 000 000 000 353 099 4;
  • 2) 0.000 000 000 000 353 099 4 × 2 = 0 + 0.000 000 000 000 706 198 8;
  • 3) 0.000 000 000 000 706 198 8 × 2 = 0 + 0.000 000 000 001 412 397 6;
  • 4) 0.000 000 000 001 412 397 6 × 2 = 0 + 0.000 000 000 002 824 795 2;
  • 5) 0.000 000 000 002 824 795 2 × 2 = 0 + 0.000 000 000 005 649 590 4;
  • 6) 0.000 000 000 005 649 590 4 × 2 = 0 + 0.000 000 000 011 299 180 8;
  • 7) 0.000 000 000 011 299 180 8 × 2 = 0 + 0.000 000 000 022 598 361 6;
  • 8) 0.000 000 000 022 598 361 6 × 2 = 0 + 0.000 000 000 045 196 723 2;
  • 9) 0.000 000 000 045 196 723 2 × 2 = 0 + 0.000 000 000 090 393 446 4;
  • 10) 0.000 000 000 090 393 446 4 × 2 = 0 + 0.000 000 000 180 786 892 8;
  • 11) 0.000 000 000 180 786 892 8 × 2 = 0 + 0.000 000 000 361 573 785 6;
  • 12) 0.000 000 000 361 573 785 6 × 2 = 0 + 0.000 000 000 723 147 571 2;
  • 13) 0.000 000 000 723 147 571 2 × 2 = 0 + 0.000 000 001 446 295 142 4;
  • 14) 0.000 000 001 446 295 142 4 × 2 = 0 + 0.000 000 002 892 590 284 8;
  • 15) 0.000 000 002 892 590 284 8 × 2 = 0 + 0.000 000 005 785 180 569 6;
  • 16) 0.000 000 005 785 180 569 6 × 2 = 0 + 0.000 000 011 570 361 139 2;
  • 17) 0.000 000 011 570 361 139 2 × 2 = 0 + 0.000 000 023 140 722 278 4;
  • 18) 0.000 000 023 140 722 278 4 × 2 = 0 + 0.000 000 046 281 444 556 8;
  • 19) 0.000 000 046 281 444 556 8 × 2 = 0 + 0.000 000 092 562 889 113 6;
  • 20) 0.000 000 092 562 889 113 6 × 2 = 0 + 0.000 000 185 125 778 227 2;
  • 21) 0.000 000 185 125 778 227 2 × 2 = 0 + 0.000 000 370 251 556 454 4;
  • 22) 0.000 000 370 251 556 454 4 × 2 = 0 + 0.000 000 740 503 112 908 8;
  • 23) 0.000 000 740 503 112 908 8 × 2 = 0 + 0.000 001 481 006 225 817 6;
  • 24) 0.000 001 481 006 225 817 6 × 2 = 0 + 0.000 002 962 012 451 635 2;
  • 25) 0.000 002 962 012 451 635 2 × 2 = 0 + 0.000 005 924 024 903 270 4;
  • 26) 0.000 005 924 024 903 270 4 × 2 = 0 + 0.000 011 848 049 806 540 8;
  • 27) 0.000 011 848 049 806 540 8 × 2 = 0 + 0.000 023 696 099 613 081 6;
  • 28) 0.000 023 696 099 613 081 6 × 2 = 0 + 0.000 047 392 199 226 163 2;
  • 29) 0.000 047 392 199 226 163 2 × 2 = 0 + 0.000 094 784 398 452 326 4;
  • 30) 0.000 094 784 398 452 326 4 × 2 = 0 + 0.000 189 568 796 904 652 8;
  • 31) 0.000 189 568 796 904 652 8 × 2 = 0 + 0.000 379 137 593 809 305 6;
  • 32) 0.000 379 137 593 809 305 6 × 2 = 0 + 0.000 758 275 187 618 611 2;
  • 33) 0.000 758 275 187 618 611 2 × 2 = 0 + 0.001 516 550 375 237 222 4;
  • 34) 0.001 516 550 375 237 222 4 × 2 = 0 + 0.003 033 100 750 474 444 8;
  • 35) 0.003 033 100 750 474 444 8 × 2 = 0 + 0.006 066 201 500 948 889 6;
  • 36) 0.006 066 201 500 948 889 6 × 2 = 0 + 0.012 132 403 001 897 779 2;
  • 37) 0.012 132 403 001 897 779 2 × 2 = 0 + 0.024 264 806 003 795 558 4;
  • 38) 0.024 264 806 003 795 558 4 × 2 = 0 + 0.048 529 612 007 591 116 8;
  • 39) 0.048 529 612 007 591 116 8 × 2 = 0 + 0.097 059 224 015 182 233 6;
  • 40) 0.097 059 224 015 182 233 6 × 2 = 0 + 0.194 118 448 030 364 467 2;
  • 41) 0.194 118 448 030 364 467 2 × 2 = 0 + 0.388 236 896 060 728 934 4;
  • 42) 0.388 236 896 060 728 934 4 × 2 = 0 + 0.776 473 792 121 457 868 8;
  • 43) 0.776 473 792 121 457 868 8 × 2 = 1 + 0.552 947 584 242 915 737 6;
  • 44) 0.552 947 584 242 915 737 6 × 2 = 1 + 0.105 895 168 485 831 475 2;
  • 45) 0.105 895 168 485 831 475 2 × 2 = 0 + 0.211 790 336 971 662 950 4;
  • 46) 0.211 790 336 971 662 950 4 × 2 = 0 + 0.423 580 673 943 325 900 8;
  • 47) 0.423 580 673 943 325 900 8 × 2 = 0 + 0.847 161 347 886 651 801 6;
  • 48) 0.847 161 347 886 651 801 6 × 2 = 1 + 0.694 322 695 773 303 603 2;
  • 49) 0.694 322 695 773 303 603 2 × 2 = 1 + 0.388 645 391 546 607 206 4;
  • 50) 0.388 645 391 546 607 206 4 × 2 = 0 + 0.777 290 783 093 214 412 8;
  • 51) 0.777 290 783 093 214 412 8 × 2 = 1 + 0.554 581 566 186 428 825 6;
  • 52) 0.554 581 566 186 428 825 6 × 2 = 1 + 0.109 163 132 372 857 651 2;
  • 53) 0.109 163 132 372 857 651 2 × 2 = 0 + 0.218 326 264 745 715 302 4;
  • 54) 0.218 326 264 745 715 302 4 × 2 = 0 + 0.436 652 529 491 430 604 8;
  • 55) 0.436 652 529 491 430 604 8 × 2 = 0 + 0.873 305 058 982 861 209 6;
  • 56) 0.873 305 058 982 861 209 6 × 2 = 1 + 0.746 610 117 965 722 419 2;
  • 57) 0.746 610 117 965 722 419 2 × 2 = 1 + 0.493 220 235 931 444 838 4;
  • 58) 0.493 220 235 931 444 838 4 × 2 = 0 + 0.986 440 471 862 889 676 8;
  • 59) 0.986 440 471 862 889 676 8 × 2 = 1 + 0.972 880 943 725 779 353 6;
  • 60) 0.972 880 943 725 779 353 6 × 2 = 1 + 0.945 761 887 451 558 707 2;
  • 61) 0.945 761 887 451 558 707 2 × 2 = 1 + 0.891 523 774 903 117 414 4;
  • 62) 0.891 523 774 903 117 414 4 × 2 = 1 + 0.783 047 549 806 234 828 8;
  • 63) 0.783 047 549 806 234 828 8 × 2 = 1 + 0.566 095 099 612 469 657 6;
  • 64) 0.566 095 099 612 469 657 6 × 2 = 1 + 0.132 190 199 224 939 315 2;
  • 65) 0.132 190 199 224 939 315 2 × 2 = 0 + 0.264 380 398 449 878 630 4;
  • 66) 0.264 380 398 449 878 630 4 × 2 = 0 + 0.528 760 796 899 757 260 8;
  • 67) 0.528 760 796 899 757 260 8 × 2 = 1 + 0.057 521 593 799 514 521 6;
  • 68) 0.057 521 593 799 514 521 6 × 2 = 0 + 0.115 043 187 599 029 043 2;
  • 69) 0.115 043 187 599 029 043 2 × 2 = 0 + 0.230 086 375 198 058 086 4;
  • 70) 0.230 086 375 198 058 086 4 × 2 = 0 + 0.460 172 750 396 116 172 8;
  • 71) 0.460 172 750 396 116 172 8 × 2 = 0 + 0.920 345 500 792 232 345 6;
  • 72) 0.920 345 500 792 232 345 6 × 2 = 1 + 0.840 691 001 584 464 691 2;
  • 73) 0.840 691 001 584 464 691 2 × 2 = 1 + 0.681 382 003 168 929 382 4;
  • 74) 0.681 382 003 168 929 382 4 × 2 = 1 + 0.362 764 006 337 858 764 8;
  • 75) 0.362 764 006 337 858 764 8 × 2 = 0 + 0.725 528 012 675 717 529 6;
  • 76) 0.725 528 012 675 717 529 6 × 2 = 1 + 0.451 056 025 351 435 059 2;
  • 77) 0.451 056 025 351 435 059 2 × 2 = 0 + 0.902 112 050 702 870 118 4;
  • 78) 0.902 112 050 702 870 118 4 × 2 = 1 + 0.804 224 101 405 740 236 8;
  • 79) 0.804 224 101 405 740 236 8 × 2 = 1 + 0.608 448 202 811 480 473 6;
  • 80) 0.608 448 202 811 480 473 6 × 2 = 1 + 0.216 896 405 622 960 947 2;
  • 81) 0.216 896 405 622 960 947 2 × 2 = 0 + 0.433 792 811 245 921 894 4;
  • 82) 0.433 792 811 245 921 894 4 × 2 = 0 + 0.867 585 622 491 843 788 8;
  • 83) 0.867 585 622 491 843 788 8 × 2 = 1 + 0.735 171 244 983 687 577 6;
  • 84) 0.735 171 244 983 687 577 6 × 2 = 1 + 0.470 342 489 967 375 155 2;
  • 85) 0.470 342 489 967 375 155 2 × 2 = 0 + 0.940 684 979 934 750 310 4;
  • 86) 0.940 684 979 934 750 310 4 × 2 = 1 + 0.881 369 959 869 500 620 8;
  • 87) 0.881 369 959 869 500 620 8 × 2 = 1 + 0.762 739 919 739 001 241 6;
  • 88) 0.762 739 919 739 001 241 6 × 2 = 1 + 0.525 479 839 478 002 483 2;
  • 89) 0.525 479 839 478 002 483 2 × 2 = 1 + 0.050 959 678 956 004 966 4;
  • 90) 0.050 959 678 956 004 966 4 × 2 = 0 + 0.101 919 357 912 009 932 8;
  • 91) 0.101 919 357 912 009 932 8 × 2 = 0 + 0.203 838 715 824 019 865 6;
  • 92) 0.203 838 715 824 019 865 6 × 2 = 0 + 0.407 677 431 648 039 731 2;
  • 93) 0.407 677 431 648 039 731 2 × 2 = 0 + 0.815 354 863 296 079 462 4;
  • 94) 0.815 354 863 296 079 462 4 × 2 = 1 + 0.630 709 726 592 158 924 8;
  • 95) 0.630 709 726 592 158 924 8 × 2 = 1 + 0.261 419 453 184 317 849 6;

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:


0.000 000 000 000 176 549 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1011 1111 0010 0001 1101 0111 0011 0111 1000 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 549 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1011 1111 0010 0001 1101 0111 0011 0111 1000 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:


0.000 000 000 000 176 549 7(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1011 1111 0010 0001 1101 0111 0011 0111 1000 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1011 1111 0010 0001 1101 0111 0011 0111 1000 011(2) × 20 =

1.1000 1101 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 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 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 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 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 0011 =

1000 1101 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 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 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 0011


Decimal number -0.000 000 000 000 176 549 7 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1101 1111 1001 0000 1110 1011 1001 1011 1100 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