-0.000 000 000 000 014 353 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 353 9(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
-0.000 000 000 000 014 353 9(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:

|-0.000 000 000 000 014 353 9| = 0.000 000 000 000 014 353 9


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: 0.000 000 000 000 014 353 9.

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) 0.000 000 000 000 014 353 9 × 2 = 0 + 0.000 000 000 000 028 707 8;
  • 2) 0.000 000 000 000 028 707 8 × 2 = 0 + 0.000 000 000 000 057 415 6;
  • 3) 0.000 000 000 000 057 415 6 × 2 = 0 + 0.000 000 000 000 114 831 2;
  • 4) 0.000 000 000 000 114 831 2 × 2 = 0 + 0.000 000 000 000 229 662 4;
  • 5) 0.000 000 000 000 229 662 4 × 2 = 0 + 0.000 000 000 000 459 324 8;
  • 6) 0.000 000 000 000 459 324 8 × 2 = 0 + 0.000 000 000 000 918 649 6;
  • 7) 0.000 000 000 000 918 649 6 × 2 = 0 + 0.000 000 000 001 837 299 2;
  • 8) 0.000 000 000 001 837 299 2 × 2 = 0 + 0.000 000 000 003 674 598 4;
  • 9) 0.000 000 000 003 674 598 4 × 2 = 0 + 0.000 000 000 007 349 196 8;
  • 10) 0.000 000 000 007 349 196 8 × 2 = 0 + 0.000 000 000 014 698 393 6;
  • 11) 0.000 000 000 014 698 393 6 × 2 = 0 + 0.000 000 000 029 396 787 2;
  • 12) 0.000 000 000 029 396 787 2 × 2 = 0 + 0.000 000 000 058 793 574 4;
  • 13) 0.000 000 000 058 793 574 4 × 2 = 0 + 0.000 000 000 117 587 148 8;
  • 14) 0.000 000 000 117 587 148 8 × 2 = 0 + 0.000 000 000 235 174 297 6;
  • 15) 0.000 000 000 235 174 297 6 × 2 = 0 + 0.000 000 000 470 348 595 2;
  • 16) 0.000 000 000 470 348 595 2 × 2 = 0 + 0.000 000 000 940 697 190 4;
  • 17) 0.000 000 000 940 697 190 4 × 2 = 0 + 0.000 000 001 881 394 380 8;
  • 18) 0.000 000 001 881 394 380 8 × 2 = 0 + 0.000 000 003 762 788 761 6;
  • 19) 0.000 000 003 762 788 761 6 × 2 = 0 + 0.000 000 007 525 577 523 2;
  • 20) 0.000 000 007 525 577 523 2 × 2 = 0 + 0.000 000 015 051 155 046 4;
  • 21) 0.000 000 015 051 155 046 4 × 2 = 0 + 0.000 000 030 102 310 092 8;
  • 22) 0.000 000 030 102 310 092 8 × 2 = 0 + 0.000 000 060 204 620 185 6;
  • 23) 0.000 000 060 204 620 185 6 × 2 = 0 + 0.000 000 120 409 240 371 2;
  • 24) 0.000 000 120 409 240 371 2 × 2 = 0 + 0.000 000 240 818 480 742 4;
  • 25) 0.000 000 240 818 480 742 4 × 2 = 0 + 0.000 000 481 636 961 484 8;
  • 26) 0.000 000 481 636 961 484 8 × 2 = 0 + 0.000 000 963 273 922 969 6;
  • 27) 0.000 000 963 273 922 969 6 × 2 = 0 + 0.000 001 926 547 845 939 2;
  • 28) 0.000 001 926 547 845 939 2 × 2 = 0 + 0.000 003 853 095 691 878 4;
  • 29) 0.000 003 853 095 691 878 4 × 2 = 0 + 0.000 007 706 191 383 756 8;
  • 30) 0.000 007 706 191 383 756 8 × 2 = 0 + 0.000 015 412 382 767 513 6;
  • 31) 0.000 015 412 382 767 513 6 × 2 = 0 + 0.000 030 824 765 535 027 2;
  • 32) 0.000 030 824 765 535 027 2 × 2 = 0 + 0.000 061 649 531 070 054 4;
  • 33) 0.000 061 649 531 070 054 4 × 2 = 0 + 0.000 123 299 062 140 108 8;
  • 34) 0.000 123 299 062 140 108 8 × 2 = 0 + 0.000 246 598 124 280 217 6;
  • 35) 0.000 246 598 124 280 217 6 × 2 = 0 + 0.000 493 196 248 560 435 2;
  • 36) 0.000 493 196 248 560 435 2 × 2 = 0 + 0.000 986 392 497 120 870 4;
  • 37) 0.000 986 392 497 120 870 4 × 2 = 0 + 0.001 972 784 994 241 740 8;
  • 38) 0.001 972 784 994 241 740 8 × 2 = 0 + 0.003 945 569 988 483 481 6;
  • 39) 0.003 945 569 988 483 481 6 × 2 = 0 + 0.007 891 139 976 966 963 2;
  • 40) 0.007 891 139 976 966 963 2 × 2 = 0 + 0.015 782 279 953 933 926 4;
  • 41) 0.015 782 279 953 933 926 4 × 2 = 0 + 0.031 564 559 907 867 852 8;
  • 42) 0.031 564 559 907 867 852 8 × 2 = 0 + 0.063 129 119 815 735 705 6;
  • 43) 0.063 129 119 815 735 705 6 × 2 = 0 + 0.126 258 239 631 471 411 2;
  • 44) 0.126 258 239 631 471 411 2 × 2 = 0 + 0.252 516 479 262 942 822 4;
  • 45) 0.252 516 479 262 942 822 4 × 2 = 0 + 0.505 032 958 525 885 644 8;
  • 46) 0.505 032 958 525 885 644 8 × 2 = 1 + 0.010 065 917 051 771 289 6;
  • 47) 0.010 065 917 051 771 289 6 × 2 = 0 + 0.020 131 834 103 542 579 2;
  • 48) 0.020 131 834 103 542 579 2 × 2 = 0 + 0.040 263 668 207 085 158 4;
  • 49) 0.040 263 668 207 085 158 4 × 2 = 0 + 0.080 527 336 414 170 316 8;
  • 50) 0.080 527 336 414 170 316 8 × 2 = 0 + 0.161 054 672 828 340 633 6;
  • 51) 0.161 054 672 828 340 633 6 × 2 = 0 + 0.322 109 345 656 681 267 2;
  • 52) 0.322 109 345 656 681 267 2 × 2 = 0 + 0.644 218 691 313 362 534 4;
  • 53) 0.644 218 691 313 362 534 4 × 2 = 1 + 0.288 437 382 626 725 068 8;
  • 54) 0.288 437 382 626 725 068 8 × 2 = 0 + 0.576 874 765 253 450 137 6;
  • 55) 0.576 874 765 253 450 137 6 × 2 = 1 + 0.153 749 530 506 900 275 2;
  • 56) 0.153 749 530 506 900 275 2 × 2 = 0 + 0.307 499 061 013 800 550 4;
  • 57) 0.307 499 061 013 800 550 4 × 2 = 0 + 0.614 998 122 027 601 100 8;
  • 58) 0.614 998 122 027 601 100 8 × 2 = 1 + 0.229 996 244 055 202 201 6;
  • 59) 0.229 996 244 055 202 201 6 × 2 = 0 + 0.459 992 488 110 404 403 2;
  • 60) 0.459 992 488 110 404 403 2 × 2 = 0 + 0.919 984 976 220 808 806 4;
  • 61) 0.919 984 976 220 808 806 4 × 2 = 1 + 0.839 969 952 441 617 612 8;
  • 62) 0.839 969 952 441 617 612 8 × 2 = 1 + 0.679 939 904 883 235 225 6;
  • 63) 0.679 939 904 883 235 225 6 × 2 = 1 + 0.359 879 809 766 470 451 2;
  • 64) 0.359 879 809 766 470 451 2 × 2 = 0 + 0.719 759 619 532 940 902 4;
  • 65) 0.719 759 619 532 940 902 4 × 2 = 1 + 0.439 519 239 065 881 804 8;
  • 66) 0.439 519 239 065 881 804 8 × 2 = 0 + 0.879 038 478 131 763 609 6;
  • 67) 0.879 038 478 131 763 609 6 × 2 = 1 + 0.758 076 956 263 527 219 2;
  • 68) 0.758 076 956 263 527 219 2 × 2 = 1 + 0.516 153 912 527 054 438 4;
  • 69) 0.516 153 912 527 054 438 4 × 2 = 1 + 0.032 307 825 054 108 876 8;
  • 70) 0.032 307 825 054 108 876 8 × 2 = 0 + 0.064 615 650 108 217 753 6;
  • 71) 0.064 615 650 108 217 753 6 × 2 = 0 + 0.129 231 300 216 435 507 2;
  • 72) 0.129 231 300 216 435 507 2 × 2 = 0 + 0.258 462 600 432 871 014 4;
  • 73) 0.258 462 600 432 871 014 4 × 2 = 0 + 0.516 925 200 865 742 028 8;
  • 74) 0.516 925 200 865 742 028 8 × 2 = 1 + 0.033 850 401 731 484 057 6;
  • 75) 0.033 850 401 731 484 057 6 × 2 = 0 + 0.067 700 803 462 968 115 2;
  • 76) 0.067 700 803 462 968 115 2 × 2 = 0 + 0.135 401 606 925 936 230 4;
  • 77) 0.135 401 606 925 936 230 4 × 2 = 0 + 0.270 803 213 851 872 460 8;
  • 78) 0.270 803 213 851 872 460 8 × 2 = 0 + 0.541 606 427 703 744 921 6;
  • 79) 0.541 606 427 703 744 921 6 × 2 = 1 + 0.083 212 855 407 489 843 2;
  • 80) 0.083 212 855 407 489 843 2 × 2 = 0 + 0.166 425 710 814 979 686 4;
  • 81) 0.166 425 710 814 979 686 4 × 2 = 0 + 0.332 851 421 629 959 372 8;
  • 82) 0.332 851 421 629 959 372 8 × 2 = 0 + 0.665 702 843 259 918 745 6;
  • 83) 0.665 702 843 259 918 745 6 × 2 = 1 + 0.331 405 686 519 837 491 2;
  • 84) 0.331 405 686 519 837 491 2 × 2 = 0 + 0.662 811 373 039 674 982 4;
  • 85) 0.662 811 373 039 674 982 4 × 2 = 1 + 0.325 622 746 079 349 964 8;
  • 86) 0.325 622 746 079 349 964 8 × 2 = 0 + 0.651 245 492 158 699 929 6;
  • 87) 0.651 245 492 158 699 929 6 × 2 = 1 + 0.302 490 984 317 399 859 2;
  • 88) 0.302 490 984 317 399 859 2 × 2 = 0 + 0.604 981 968 634 799 718 4;
  • 89) 0.604 981 968 634 799 718 4 × 2 = 1 + 0.209 963 937 269 599 436 8;
  • 90) 0.209 963 937 269 599 436 8 × 2 = 0 + 0.419 927 874 539 198 873 6;
  • 91) 0.419 927 874 539 198 873 6 × 2 = 0 + 0.839 855 749 078 397 747 2;
  • 92) 0.839 855 749 078 397 747 2 × 2 = 1 + 0.679 711 498 156 795 494 4;
  • 93) 0.679 711 498 156 795 494 4 × 2 = 1 + 0.359 422 996 313 590 988 8;
  • 94) 0.359 422 996 313 590 988 8 × 2 = 0 + 0.718 845 992 627 181 977 6;
  • 95) 0.718 845 992 627 181 977 6 × 2 = 1 + 0.437 691 985 254 363 955 2;
  • 96) 0.437 691 985 254 363 955 2 × 2 = 0 + 0.875 383 970 508 727 910 4;
  • 97) 0.875 383 970 508 727 910 4 × 2 = 1 + 0.750 767 941 017 455 820 8;
  • 98) 0.750 767 941 017 455 820 8 × 2 = 1 + 0.501 535 882 034 911 641 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:


0.000 000 000 000 014 353 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0100 1110 1011 1000 0100 0010 0010 1010 1001 1010 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 353 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0100 1110 1011 1000 0100 0010 0010 1010 1001 1010 11(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:


0.000 000 000 000 014 353 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0100 1110 1011 1000 0100 0010 0010 1010 1001 1010 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1010 0100 1110 1011 1000 0100 0010 0010 1010 1001 1010 11(2) × 20 =

1.0000 0010 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011(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 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011


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 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011 =

0000 0010 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011


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 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011


Decimal number -0.000 000 000 000 014 353 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1001 0011 1010 1110 0001 0000 1000 1010 1010 0110 1011

» Convert decimal -0.000 000 000 000 014 359 9 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