-0.000 000 000 000 014 373 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 373 8(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
-0.000 000 000 000 014 373 8(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:

|-0.000 000 000 000 014 373 8| = 0.000 000 000 000 014 373 8


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: 0.000 000 000 000 014 373 8.

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) 0.000 000 000 000 014 373 8 × 2 = 0 + 0.000 000 000 000 028 747 6;
  • 2) 0.000 000 000 000 028 747 6 × 2 = 0 + 0.000 000 000 000 057 495 2;
  • 3) 0.000 000 000 000 057 495 2 × 2 = 0 + 0.000 000 000 000 114 990 4;
  • 4) 0.000 000 000 000 114 990 4 × 2 = 0 + 0.000 000 000 000 229 980 8;
  • 5) 0.000 000 000 000 229 980 8 × 2 = 0 + 0.000 000 000 000 459 961 6;
  • 6) 0.000 000 000 000 459 961 6 × 2 = 0 + 0.000 000 000 000 919 923 2;
  • 7) 0.000 000 000 000 919 923 2 × 2 = 0 + 0.000 000 000 001 839 846 4;
  • 8) 0.000 000 000 001 839 846 4 × 2 = 0 + 0.000 000 000 003 679 692 8;
  • 9) 0.000 000 000 003 679 692 8 × 2 = 0 + 0.000 000 000 007 359 385 6;
  • 10) 0.000 000 000 007 359 385 6 × 2 = 0 + 0.000 000 000 014 718 771 2;
  • 11) 0.000 000 000 014 718 771 2 × 2 = 0 + 0.000 000 000 029 437 542 4;
  • 12) 0.000 000 000 029 437 542 4 × 2 = 0 + 0.000 000 000 058 875 084 8;
  • 13) 0.000 000 000 058 875 084 8 × 2 = 0 + 0.000 000 000 117 750 169 6;
  • 14) 0.000 000 000 117 750 169 6 × 2 = 0 + 0.000 000 000 235 500 339 2;
  • 15) 0.000 000 000 235 500 339 2 × 2 = 0 + 0.000 000 000 471 000 678 4;
  • 16) 0.000 000 000 471 000 678 4 × 2 = 0 + 0.000 000 000 942 001 356 8;
  • 17) 0.000 000 000 942 001 356 8 × 2 = 0 + 0.000 000 001 884 002 713 6;
  • 18) 0.000 000 001 884 002 713 6 × 2 = 0 + 0.000 000 003 768 005 427 2;
  • 19) 0.000 000 003 768 005 427 2 × 2 = 0 + 0.000 000 007 536 010 854 4;
  • 20) 0.000 000 007 536 010 854 4 × 2 = 0 + 0.000 000 015 072 021 708 8;
  • 21) 0.000 000 015 072 021 708 8 × 2 = 0 + 0.000 000 030 144 043 417 6;
  • 22) 0.000 000 030 144 043 417 6 × 2 = 0 + 0.000 000 060 288 086 835 2;
  • 23) 0.000 000 060 288 086 835 2 × 2 = 0 + 0.000 000 120 576 173 670 4;
  • 24) 0.000 000 120 576 173 670 4 × 2 = 0 + 0.000 000 241 152 347 340 8;
  • 25) 0.000 000 241 152 347 340 8 × 2 = 0 + 0.000 000 482 304 694 681 6;
  • 26) 0.000 000 482 304 694 681 6 × 2 = 0 + 0.000 000 964 609 389 363 2;
  • 27) 0.000 000 964 609 389 363 2 × 2 = 0 + 0.000 001 929 218 778 726 4;
  • 28) 0.000 001 929 218 778 726 4 × 2 = 0 + 0.000 003 858 437 557 452 8;
  • 29) 0.000 003 858 437 557 452 8 × 2 = 0 + 0.000 007 716 875 114 905 6;
  • 30) 0.000 007 716 875 114 905 6 × 2 = 0 + 0.000 015 433 750 229 811 2;
  • 31) 0.000 015 433 750 229 811 2 × 2 = 0 + 0.000 030 867 500 459 622 4;
  • 32) 0.000 030 867 500 459 622 4 × 2 = 0 + 0.000 061 735 000 919 244 8;
  • 33) 0.000 061 735 000 919 244 8 × 2 = 0 + 0.000 123 470 001 838 489 6;
  • 34) 0.000 123 470 001 838 489 6 × 2 = 0 + 0.000 246 940 003 676 979 2;
  • 35) 0.000 246 940 003 676 979 2 × 2 = 0 + 0.000 493 880 007 353 958 4;
  • 36) 0.000 493 880 007 353 958 4 × 2 = 0 + 0.000 987 760 014 707 916 8;
  • 37) 0.000 987 760 014 707 916 8 × 2 = 0 + 0.001 975 520 029 415 833 6;
  • 38) 0.001 975 520 029 415 833 6 × 2 = 0 + 0.003 951 040 058 831 667 2;
  • 39) 0.003 951 040 058 831 667 2 × 2 = 0 + 0.007 902 080 117 663 334 4;
  • 40) 0.007 902 080 117 663 334 4 × 2 = 0 + 0.015 804 160 235 326 668 8;
  • 41) 0.015 804 160 235 326 668 8 × 2 = 0 + 0.031 608 320 470 653 337 6;
  • 42) 0.031 608 320 470 653 337 6 × 2 = 0 + 0.063 216 640 941 306 675 2;
  • 43) 0.063 216 640 941 306 675 2 × 2 = 0 + 0.126 433 281 882 613 350 4;
  • 44) 0.126 433 281 882 613 350 4 × 2 = 0 + 0.252 866 563 765 226 700 8;
  • 45) 0.252 866 563 765 226 700 8 × 2 = 0 + 0.505 733 127 530 453 401 6;
  • 46) 0.505 733 127 530 453 401 6 × 2 = 1 + 0.011 466 255 060 906 803 2;
  • 47) 0.011 466 255 060 906 803 2 × 2 = 0 + 0.022 932 510 121 813 606 4;
  • 48) 0.022 932 510 121 813 606 4 × 2 = 0 + 0.045 865 020 243 627 212 8;
  • 49) 0.045 865 020 243 627 212 8 × 2 = 0 + 0.091 730 040 487 254 425 6;
  • 50) 0.091 730 040 487 254 425 6 × 2 = 0 + 0.183 460 080 974 508 851 2;
  • 51) 0.183 460 080 974 508 851 2 × 2 = 0 + 0.366 920 161 949 017 702 4;
  • 52) 0.366 920 161 949 017 702 4 × 2 = 0 + 0.733 840 323 898 035 404 8;
  • 53) 0.733 840 323 898 035 404 8 × 2 = 1 + 0.467 680 647 796 070 809 6;
  • 54) 0.467 680 647 796 070 809 6 × 2 = 0 + 0.935 361 295 592 141 619 2;
  • 55) 0.935 361 295 592 141 619 2 × 2 = 1 + 0.870 722 591 184 283 238 4;
  • 56) 0.870 722 591 184 283 238 4 × 2 = 1 + 0.741 445 182 368 566 476 8;
  • 57) 0.741 445 182 368 566 476 8 × 2 = 1 + 0.482 890 364 737 132 953 6;
  • 58) 0.482 890 364 737 132 953 6 × 2 = 0 + 0.965 780 729 474 265 907 2;
  • 59) 0.965 780 729 474 265 907 2 × 2 = 1 + 0.931 561 458 948 531 814 4;
  • 60) 0.931 561 458 948 531 814 4 × 2 = 1 + 0.863 122 917 897 063 628 8;
  • 61) 0.863 122 917 897 063 628 8 × 2 = 1 + 0.726 245 835 794 127 257 6;
  • 62) 0.726 245 835 794 127 257 6 × 2 = 1 + 0.452 491 671 588 254 515 2;
  • 63) 0.452 491 671 588 254 515 2 × 2 = 0 + 0.904 983 343 176 509 030 4;
  • 64) 0.904 983 343 176 509 030 4 × 2 = 1 + 0.809 966 686 353 018 060 8;
  • 65) 0.809 966 686 353 018 060 8 × 2 = 1 + 0.619 933 372 706 036 121 6;
  • 66) 0.619 933 372 706 036 121 6 × 2 = 1 + 0.239 866 745 412 072 243 2;
  • 67) 0.239 866 745 412 072 243 2 × 2 = 0 + 0.479 733 490 824 144 486 4;
  • 68) 0.479 733 490 824 144 486 4 × 2 = 0 + 0.959 466 981 648 288 972 8;
  • 69) 0.959 466 981 648 288 972 8 × 2 = 1 + 0.918 933 963 296 577 945 6;
  • 70) 0.918 933 963 296 577 945 6 × 2 = 1 + 0.837 867 926 593 155 891 2;
  • 71) 0.837 867 926 593 155 891 2 × 2 = 1 + 0.675 735 853 186 311 782 4;
  • 72) 0.675 735 853 186 311 782 4 × 2 = 1 + 0.351 471 706 372 623 564 8;
  • 73) 0.351 471 706 372 623 564 8 × 2 = 0 + 0.702 943 412 745 247 129 6;
  • 74) 0.702 943 412 745 247 129 6 × 2 = 1 + 0.405 886 825 490 494 259 2;
  • 75) 0.405 886 825 490 494 259 2 × 2 = 0 + 0.811 773 650 980 988 518 4;
  • 76) 0.811 773 650 980 988 518 4 × 2 = 1 + 0.623 547 301 961 977 036 8;
  • 77) 0.623 547 301 961 977 036 8 × 2 = 1 + 0.247 094 603 923 954 073 6;
  • 78) 0.247 094 603 923 954 073 6 × 2 = 0 + 0.494 189 207 847 908 147 2;
  • 79) 0.494 189 207 847 908 147 2 × 2 = 0 + 0.988 378 415 695 816 294 4;
  • 80) 0.988 378 415 695 816 294 4 × 2 = 1 + 0.976 756 831 391 632 588 8;
  • 81) 0.976 756 831 391 632 588 8 × 2 = 1 + 0.953 513 662 783 265 177 6;
  • 82) 0.953 513 662 783 265 177 6 × 2 = 1 + 0.907 027 325 566 530 355 2;
  • 83) 0.907 027 325 566 530 355 2 × 2 = 1 + 0.814 054 651 133 060 710 4;
  • 84) 0.814 054 651 133 060 710 4 × 2 = 1 + 0.628 109 302 266 121 420 8;
  • 85) 0.628 109 302 266 121 420 8 × 2 = 1 + 0.256 218 604 532 242 841 6;
  • 86) 0.256 218 604 532 242 841 6 × 2 = 0 + 0.512 437 209 064 485 683 2;
  • 87) 0.512 437 209 064 485 683 2 × 2 = 1 + 0.024 874 418 128 971 366 4;
  • 88) 0.024 874 418 128 971 366 4 × 2 = 0 + 0.049 748 836 257 942 732 8;
  • 89) 0.049 748 836 257 942 732 8 × 2 = 0 + 0.099 497 672 515 885 465 6;
  • 90) 0.099 497 672 515 885 465 6 × 2 = 0 + 0.198 995 345 031 770 931 2;
  • 91) 0.198 995 345 031 770 931 2 × 2 = 0 + 0.397 990 690 063 541 862 4;
  • 92) 0.397 990 690 063 541 862 4 × 2 = 0 + 0.795 981 380 127 083 724 8;
  • 93) 0.795 981 380 127 083 724 8 × 2 = 1 + 0.591 962 760 254 167 449 6;
  • 94) 0.591 962 760 254 167 449 6 × 2 = 1 + 0.183 925 520 508 334 899 2;
  • 95) 0.183 925 520 508 334 899 2 × 2 = 0 + 0.367 851 041 016 669 798 4;
  • 96) 0.367 851 041 016 669 798 4 × 2 = 0 + 0.735 702 082 033 339 596 8;
  • 97) 0.735 702 082 033 339 596 8 × 2 = 1 + 0.471 404 164 066 679 193 6;
  • 98) 0.471 404 164 066 679 193 6 × 2 = 0 + 0.942 808 328 133 358 387 2;

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:


0.000 000 000 000 014 373 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 1100 1111 0101 1001 1111 1010 0000 1100 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 373 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 1100 1111 0101 1001 1111 1010 0000 1100 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:


0.000 000 000 000 014 373 8(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 1100 1111 0101 1001 1111 1010 0000 1100 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 1011 1101 1100 1111 0101 1001 1111 1010 0000 1100 10(2) × 20 =

1.0000 0010 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010(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 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010


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 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010 =

0000 0010 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010


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 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010


Decimal number -0.000 000 000 000 014 373 8 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1110 1111 0111 0011 1101 0110 0111 1110 1000 0011 0010

» Convert decimal -0.000 000 000 000 014 374 2 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