-0.000 000 000 000 014 357 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 357 6(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
-0.000 000 000 000 014 357 6(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:

|-0.000 000 000 000 014 357 6| = 0.000 000 000 000 014 357 6


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: 0.000 000 000 000 014 357 6.

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) 0.000 000 000 000 014 357 6 × 2 = 0 + 0.000 000 000 000 028 715 2;
  • 2) 0.000 000 000 000 028 715 2 × 2 = 0 + 0.000 000 000 000 057 430 4;
  • 3) 0.000 000 000 000 057 430 4 × 2 = 0 + 0.000 000 000 000 114 860 8;
  • 4) 0.000 000 000 000 114 860 8 × 2 = 0 + 0.000 000 000 000 229 721 6;
  • 5) 0.000 000 000 000 229 721 6 × 2 = 0 + 0.000 000 000 000 459 443 2;
  • 6) 0.000 000 000 000 459 443 2 × 2 = 0 + 0.000 000 000 000 918 886 4;
  • 7) 0.000 000 000 000 918 886 4 × 2 = 0 + 0.000 000 000 001 837 772 8;
  • 8) 0.000 000 000 001 837 772 8 × 2 = 0 + 0.000 000 000 003 675 545 6;
  • 9) 0.000 000 000 003 675 545 6 × 2 = 0 + 0.000 000 000 007 351 091 2;
  • 10) 0.000 000 000 007 351 091 2 × 2 = 0 + 0.000 000 000 014 702 182 4;
  • 11) 0.000 000 000 014 702 182 4 × 2 = 0 + 0.000 000 000 029 404 364 8;
  • 12) 0.000 000 000 029 404 364 8 × 2 = 0 + 0.000 000 000 058 808 729 6;
  • 13) 0.000 000 000 058 808 729 6 × 2 = 0 + 0.000 000 000 117 617 459 2;
  • 14) 0.000 000 000 117 617 459 2 × 2 = 0 + 0.000 000 000 235 234 918 4;
  • 15) 0.000 000 000 235 234 918 4 × 2 = 0 + 0.000 000 000 470 469 836 8;
  • 16) 0.000 000 000 470 469 836 8 × 2 = 0 + 0.000 000 000 940 939 673 6;
  • 17) 0.000 000 000 940 939 673 6 × 2 = 0 + 0.000 000 001 881 879 347 2;
  • 18) 0.000 000 001 881 879 347 2 × 2 = 0 + 0.000 000 003 763 758 694 4;
  • 19) 0.000 000 003 763 758 694 4 × 2 = 0 + 0.000 000 007 527 517 388 8;
  • 20) 0.000 000 007 527 517 388 8 × 2 = 0 + 0.000 000 015 055 034 777 6;
  • 21) 0.000 000 015 055 034 777 6 × 2 = 0 + 0.000 000 030 110 069 555 2;
  • 22) 0.000 000 030 110 069 555 2 × 2 = 0 + 0.000 000 060 220 139 110 4;
  • 23) 0.000 000 060 220 139 110 4 × 2 = 0 + 0.000 000 120 440 278 220 8;
  • 24) 0.000 000 120 440 278 220 8 × 2 = 0 + 0.000 000 240 880 556 441 6;
  • 25) 0.000 000 240 880 556 441 6 × 2 = 0 + 0.000 000 481 761 112 883 2;
  • 26) 0.000 000 481 761 112 883 2 × 2 = 0 + 0.000 000 963 522 225 766 4;
  • 27) 0.000 000 963 522 225 766 4 × 2 = 0 + 0.000 001 927 044 451 532 8;
  • 28) 0.000 001 927 044 451 532 8 × 2 = 0 + 0.000 003 854 088 903 065 6;
  • 29) 0.000 003 854 088 903 065 6 × 2 = 0 + 0.000 007 708 177 806 131 2;
  • 30) 0.000 007 708 177 806 131 2 × 2 = 0 + 0.000 015 416 355 612 262 4;
  • 31) 0.000 015 416 355 612 262 4 × 2 = 0 + 0.000 030 832 711 224 524 8;
  • 32) 0.000 030 832 711 224 524 8 × 2 = 0 + 0.000 061 665 422 449 049 6;
  • 33) 0.000 061 665 422 449 049 6 × 2 = 0 + 0.000 123 330 844 898 099 2;
  • 34) 0.000 123 330 844 898 099 2 × 2 = 0 + 0.000 246 661 689 796 198 4;
  • 35) 0.000 246 661 689 796 198 4 × 2 = 0 + 0.000 493 323 379 592 396 8;
  • 36) 0.000 493 323 379 592 396 8 × 2 = 0 + 0.000 986 646 759 184 793 6;
  • 37) 0.000 986 646 759 184 793 6 × 2 = 0 + 0.001 973 293 518 369 587 2;
  • 38) 0.001 973 293 518 369 587 2 × 2 = 0 + 0.003 946 587 036 739 174 4;
  • 39) 0.003 946 587 036 739 174 4 × 2 = 0 + 0.007 893 174 073 478 348 8;
  • 40) 0.007 893 174 073 478 348 8 × 2 = 0 + 0.015 786 348 146 956 697 6;
  • 41) 0.015 786 348 146 956 697 6 × 2 = 0 + 0.031 572 696 293 913 395 2;
  • 42) 0.031 572 696 293 913 395 2 × 2 = 0 + 0.063 145 392 587 826 790 4;
  • 43) 0.063 145 392 587 826 790 4 × 2 = 0 + 0.126 290 785 175 653 580 8;
  • 44) 0.126 290 785 175 653 580 8 × 2 = 0 + 0.252 581 570 351 307 161 6;
  • 45) 0.252 581 570 351 307 161 6 × 2 = 0 + 0.505 163 140 702 614 323 2;
  • 46) 0.505 163 140 702 614 323 2 × 2 = 1 + 0.010 326 281 405 228 646 4;
  • 47) 0.010 326 281 405 228 646 4 × 2 = 0 + 0.020 652 562 810 457 292 8;
  • 48) 0.020 652 562 810 457 292 8 × 2 = 0 + 0.041 305 125 620 914 585 6;
  • 49) 0.041 305 125 620 914 585 6 × 2 = 0 + 0.082 610 251 241 829 171 2;
  • 50) 0.082 610 251 241 829 171 2 × 2 = 0 + 0.165 220 502 483 658 342 4;
  • 51) 0.165 220 502 483 658 342 4 × 2 = 0 + 0.330 441 004 967 316 684 8;
  • 52) 0.330 441 004 967 316 684 8 × 2 = 0 + 0.660 882 009 934 633 369 6;
  • 53) 0.660 882 009 934 633 369 6 × 2 = 1 + 0.321 764 019 869 266 739 2;
  • 54) 0.321 764 019 869 266 739 2 × 2 = 0 + 0.643 528 039 738 533 478 4;
  • 55) 0.643 528 039 738 533 478 4 × 2 = 1 + 0.287 056 079 477 066 956 8;
  • 56) 0.287 056 079 477 066 956 8 × 2 = 0 + 0.574 112 158 954 133 913 6;
  • 57) 0.574 112 158 954 133 913 6 × 2 = 1 + 0.148 224 317 908 267 827 2;
  • 58) 0.148 224 317 908 267 827 2 × 2 = 0 + 0.296 448 635 816 535 654 4;
  • 59) 0.296 448 635 816 535 654 4 × 2 = 0 + 0.592 897 271 633 071 308 8;
  • 60) 0.592 897 271 633 071 308 8 × 2 = 1 + 0.185 794 543 266 142 617 6;
  • 61) 0.185 794 543 266 142 617 6 × 2 = 0 + 0.371 589 086 532 285 235 2;
  • 62) 0.371 589 086 532 285 235 2 × 2 = 0 + 0.743 178 173 064 570 470 4;
  • 63) 0.743 178 173 064 570 470 4 × 2 = 1 + 0.486 356 346 129 140 940 8;
  • 64) 0.486 356 346 129 140 940 8 × 2 = 0 + 0.972 712 692 258 281 881 6;
  • 65) 0.972 712 692 258 281 881 6 × 2 = 1 + 0.945 425 384 516 563 763 2;
  • 66) 0.945 425 384 516 563 763 2 × 2 = 1 + 0.890 850 769 033 127 526 4;
  • 67) 0.890 850 769 033 127 526 4 × 2 = 1 + 0.781 701 538 066 255 052 8;
  • 68) 0.781 701 538 066 255 052 8 × 2 = 1 + 0.563 403 076 132 510 105 6;
  • 69) 0.563 403 076 132 510 105 6 × 2 = 1 + 0.126 806 152 265 020 211 2;
  • 70) 0.126 806 152 265 020 211 2 × 2 = 0 + 0.253 612 304 530 040 422 4;
  • 71) 0.253 612 304 530 040 422 4 × 2 = 0 + 0.507 224 609 060 080 844 8;
  • 72) 0.507 224 609 060 080 844 8 × 2 = 1 + 0.014 449 218 120 161 689 6;
  • 73) 0.014 449 218 120 161 689 6 × 2 = 0 + 0.028 898 436 240 323 379 2;
  • 74) 0.028 898 436 240 323 379 2 × 2 = 0 + 0.057 796 872 480 646 758 4;
  • 75) 0.057 796 872 480 646 758 4 × 2 = 0 + 0.115 593 744 961 293 516 8;
  • 76) 0.115 593 744 961 293 516 8 × 2 = 0 + 0.231 187 489 922 587 033 6;
  • 77) 0.231 187 489 922 587 033 6 × 2 = 0 + 0.462 374 979 845 174 067 2;
  • 78) 0.462 374 979 845 174 067 2 × 2 = 0 + 0.924 749 959 690 348 134 4;
  • 79) 0.924 749 959 690 348 134 4 × 2 = 1 + 0.849 499 919 380 696 268 8;
  • 80) 0.849 499 919 380 696 268 8 × 2 = 1 + 0.698 999 838 761 392 537 6;
  • 81) 0.698 999 838 761 392 537 6 × 2 = 1 + 0.397 999 677 522 785 075 2;
  • 82) 0.397 999 677 522 785 075 2 × 2 = 0 + 0.795 999 355 045 570 150 4;
  • 83) 0.795 999 355 045 570 150 4 × 2 = 1 + 0.591 998 710 091 140 300 8;
  • 84) 0.591 998 710 091 140 300 8 × 2 = 1 + 0.183 997 420 182 280 601 6;
  • 85) 0.183 997 420 182 280 601 6 × 2 = 0 + 0.367 994 840 364 561 203 2;
  • 86) 0.367 994 840 364 561 203 2 × 2 = 0 + 0.735 989 680 729 122 406 4;
  • 87) 0.735 989 680 729 122 406 4 × 2 = 1 + 0.471 979 361 458 244 812 8;
  • 88) 0.471 979 361 458 244 812 8 × 2 = 0 + 0.943 958 722 916 489 625 6;
  • 89) 0.943 958 722 916 489 625 6 × 2 = 1 + 0.887 917 445 832 979 251 2;
  • 90) 0.887 917 445 832 979 251 2 × 2 = 1 + 0.775 834 891 665 958 502 4;
  • 91) 0.775 834 891 665 958 502 4 × 2 = 1 + 0.551 669 783 331 917 004 8;
  • 92) 0.551 669 783 331 917 004 8 × 2 = 1 + 0.103 339 566 663 834 009 6;
  • 93) 0.103 339 566 663 834 009 6 × 2 = 0 + 0.206 679 133 327 668 019 2;
  • 94) 0.206 679 133 327 668 019 2 × 2 = 0 + 0.413 358 266 655 336 038 4;
  • 95) 0.413 358 266 655 336 038 4 × 2 = 0 + 0.826 716 533 310 672 076 8;
  • 96) 0.826 716 533 310 672 076 8 × 2 = 1 + 0.653 433 066 621 344 153 6;
  • 97) 0.653 433 066 621 344 153 6 × 2 = 1 + 0.306 866 133 242 688 307 2;
  • 98) 0.306 866 133 242 688 307 2 × 2 = 0 + 0.613 732 266 485 376 614 4;

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:


0.000 000 000 000 014 357 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1001 0010 1111 1001 0000 0011 1011 0010 1111 0001 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 357 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1001 0010 1111 1001 0000 0011 1011 0010 1111 0001 10(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 46 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 014 357 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1001 0010 1111 1001 0000 0011 1011 0010 1111 0001 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 1001 0010 1111 1001 0000 0011 1011 0010 1111 0001 10(2) × 20 =

1.0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110(2) × 2-46


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): -46

Mantissa (not normalized):
1.0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-46 + 2(11-1) - 1 =

(-46 + 1 023)(10) =

977(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;
  • 977 ÷ 2 = 488 + 1;
  • 488 ÷ 2 = 244 + 0;
  • 244 ÷ 2 = 122 + 0;
  • 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) =

977(10) =

011 1101 0001(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. 0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110 =

0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110


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 0001

Mantissa (52 bits) =
0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110


Decimal number -0.000 000 000 000 014 357 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1010 0100 1011 1110 0100 0000 1110 1100 1011 1100 0110

» Convert decimal -0.000 000 000 000 014 362 1 to 64 bit double precision IEEE 754 binary floating point representation standard

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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