-0.000 000 000 000 014 387 6 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 387 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
-0.000 000 000 000 014 387 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:

|-0.000 000 000 000 014 387 6| = 0.000 000 000 000 014 387 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: 0.000 000 000 000 014 387 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) 0.000 000 000 000 014 387 6 × 2 = 0 + 0.000 000 000 000 028 775 2;
  • 2) 0.000 000 000 000 028 775 2 × 2 = 0 + 0.000 000 000 000 057 550 4;
  • 3) 0.000 000 000 000 057 550 4 × 2 = 0 + 0.000 000 000 000 115 100 8;
  • 4) 0.000 000 000 000 115 100 8 × 2 = 0 + 0.000 000 000 000 230 201 6;
  • 5) 0.000 000 000 000 230 201 6 × 2 = 0 + 0.000 000 000 000 460 403 2;
  • 6) 0.000 000 000 000 460 403 2 × 2 = 0 + 0.000 000 000 000 920 806 4;
  • 7) 0.000 000 000 000 920 806 4 × 2 = 0 + 0.000 000 000 001 841 612 8;
  • 8) 0.000 000 000 001 841 612 8 × 2 = 0 + 0.000 000 000 003 683 225 6;
  • 9) 0.000 000 000 003 683 225 6 × 2 = 0 + 0.000 000 000 007 366 451 2;
  • 10) 0.000 000 000 007 366 451 2 × 2 = 0 + 0.000 000 000 014 732 902 4;
  • 11) 0.000 000 000 014 732 902 4 × 2 = 0 + 0.000 000 000 029 465 804 8;
  • 12) 0.000 000 000 029 465 804 8 × 2 = 0 + 0.000 000 000 058 931 609 6;
  • 13) 0.000 000 000 058 931 609 6 × 2 = 0 + 0.000 000 000 117 863 219 2;
  • 14) 0.000 000 000 117 863 219 2 × 2 = 0 + 0.000 000 000 235 726 438 4;
  • 15) 0.000 000 000 235 726 438 4 × 2 = 0 + 0.000 000 000 471 452 876 8;
  • 16) 0.000 000 000 471 452 876 8 × 2 = 0 + 0.000 000 000 942 905 753 6;
  • 17) 0.000 000 000 942 905 753 6 × 2 = 0 + 0.000 000 001 885 811 507 2;
  • 18) 0.000 000 001 885 811 507 2 × 2 = 0 + 0.000 000 003 771 623 014 4;
  • 19) 0.000 000 003 771 623 014 4 × 2 = 0 + 0.000 000 007 543 246 028 8;
  • 20) 0.000 000 007 543 246 028 8 × 2 = 0 + 0.000 000 015 086 492 057 6;
  • 21) 0.000 000 015 086 492 057 6 × 2 = 0 + 0.000 000 030 172 984 115 2;
  • 22) 0.000 000 030 172 984 115 2 × 2 = 0 + 0.000 000 060 345 968 230 4;
  • 23) 0.000 000 060 345 968 230 4 × 2 = 0 + 0.000 000 120 691 936 460 8;
  • 24) 0.000 000 120 691 936 460 8 × 2 = 0 + 0.000 000 241 383 872 921 6;
  • 25) 0.000 000 241 383 872 921 6 × 2 = 0 + 0.000 000 482 767 745 843 2;
  • 26) 0.000 000 482 767 745 843 2 × 2 = 0 + 0.000 000 965 535 491 686 4;
  • 27) 0.000 000 965 535 491 686 4 × 2 = 0 + 0.000 001 931 070 983 372 8;
  • 28) 0.000 001 931 070 983 372 8 × 2 = 0 + 0.000 003 862 141 966 745 6;
  • 29) 0.000 003 862 141 966 745 6 × 2 = 0 + 0.000 007 724 283 933 491 2;
  • 30) 0.000 007 724 283 933 491 2 × 2 = 0 + 0.000 015 448 567 866 982 4;
  • 31) 0.000 015 448 567 866 982 4 × 2 = 0 + 0.000 030 897 135 733 964 8;
  • 32) 0.000 030 897 135 733 964 8 × 2 = 0 + 0.000 061 794 271 467 929 6;
  • 33) 0.000 061 794 271 467 929 6 × 2 = 0 + 0.000 123 588 542 935 859 2;
  • 34) 0.000 123 588 542 935 859 2 × 2 = 0 + 0.000 247 177 085 871 718 4;
  • 35) 0.000 247 177 085 871 718 4 × 2 = 0 + 0.000 494 354 171 743 436 8;
  • 36) 0.000 494 354 171 743 436 8 × 2 = 0 + 0.000 988 708 343 486 873 6;
  • 37) 0.000 988 708 343 486 873 6 × 2 = 0 + 0.001 977 416 686 973 747 2;
  • 38) 0.001 977 416 686 973 747 2 × 2 = 0 + 0.003 954 833 373 947 494 4;
  • 39) 0.003 954 833 373 947 494 4 × 2 = 0 + 0.007 909 666 747 894 988 8;
  • 40) 0.007 909 666 747 894 988 8 × 2 = 0 + 0.015 819 333 495 789 977 6;
  • 41) 0.015 819 333 495 789 977 6 × 2 = 0 + 0.031 638 666 991 579 955 2;
  • 42) 0.031 638 666 991 579 955 2 × 2 = 0 + 0.063 277 333 983 159 910 4;
  • 43) 0.063 277 333 983 159 910 4 × 2 = 0 + 0.126 554 667 966 319 820 8;
  • 44) 0.126 554 667 966 319 820 8 × 2 = 0 + 0.253 109 335 932 639 641 6;
  • 45) 0.253 109 335 932 639 641 6 × 2 = 0 + 0.506 218 671 865 279 283 2;
  • 46) 0.506 218 671 865 279 283 2 × 2 = 1 + 0.012 437 343 730 558 566 4;
  • 47) 0.012 437 343 730 558 566 4 × 2 = 0 + 0.024 874 687 461 117 132 8;
  • 48) 0.024 874 687 461 117 132 8 × 2 = 0 + 0.049 749 374 922 234 265 6;
  • 49) 0.049 749 374 922 234 265 6 × 2 = 0 + 0.099 498 749 844 468 531 2;
  • 50) 0.099 498 749 844 468 531 2 × 2 = 0 + 0.198 997 499 688 937 062 4;
  • 51) 0.198 997 499 688 937 062 4 × 2 = 0 + 0.397 994 999 377 874 124 8;
  • 52) 0.397 994 999 377 874 124 8 × 2 = 0 + 0.795 989 998 755 748 249 6;
  • 53) 0.795 989 998 755 748 249 6 × 2 = 1 + 0.591 979 997 511 496 499 2;
  • 54) 0.591 979 997 511 496 499 2 × 2 = 1 + 0.183 959 995 022 992 998 4;
  • 55) 0.183 959 995 022 992 998 4 × 2 = 0 + 0.367 919 990 045 985 996 8;
  • 56) 0.367 919 990 045 985 996 8 × 2 = 0 + 0.735 839 980 091 971 993 6;
  • 57) 0.735 839 980 091 971 993 6 × 2 = 1 + 0.471 679 960 183 943 987 2;
  • 58) 0.471 679 960 183 943 987 2 × 2 = 0 + 0.943 359 920 367 887 974 4;
  • 59) 0.943 359 920 367 887 974 4 × 2 = 1 + 0.886 719 840 735 775 948 8;
  • 60) 0.886 719 840 735 775 948 8 × 2 = 1 + 0.773 439 681 471 551 897 6;
  • 61) 0.773 439 681 471 551 897 6 × 2 = 1 + 0.546 879 362 943 103 795 2;
  • 62) 0.546 879 362 943 103 795 2 × 2 = 1 + 0.093 758 725 886 207 590 4;
  • 63) 0.093 758 725 886 207 590 4 × 2 = 0 + 0.187 517 451 772 415 180 8;
  • 64) 0.187 517 451 772 415 180 8 × 2 = 0 + 0.375 034 903 544 830 361 6;
  • 65) 0.375 034 903 544 830 361 6 × 2 = 0 + 0.750 069 807 089 660 723 2;
  • 66) 0.750 069 807 089 660 723 2 × 2 = 1 + 0.500 139 614 179 321 446 4;
  • 67) 0.500 139 614 179 321 446 4 × 2 = 1 + 0.000 279 228 358 642 892 8;
  • 68) 0.000 279 228 358 642 892 8 × 2 = 0 + 0.000 558 456 717 285 785 6;
  • 69) 0.000 558 456 717 285 785 6 × 2 = 0 + 0.001 116 913 434 571 571 2;
  • 70) 0.001 116 913 434 571 571 2 × 2 = 0 + 0.002 233 826 869 143 142 4;
  • 71) 0.002 233 826 869 143 142 4 × 2 = 0 + 0.004 467 653 738 286 284 8;
  • 72) 0.004 467 653 738 286 284 8 × 2 = 0 + 0.008 935 307 476 572 569 6;
  • 73) 0.008 935 307 476 572 569 6 × 2 = 0 + 0.017 870 614 953 145 139 2;
  • 74) 0.017 870 614 953 145 139 2 × 2 = 0 + 0.035 741 229 906 290 278 4;
  • 75) 0.035 741 229 906 290 278 4 × 2 = 0 + 0.071 482 459 812 580 556 8;
  • 76) 0.071 482 459 812 580 556 8 × 2 = 0 + 0.142 964 919 625 161 113 6;
  • 77) 0.142 964 919 625 161 113 6 × 2 = 0 + 0.285 929 839 250 322 227 2;
  • 78) 0.285 929 839 250 322 227 2 × 2 = 0 + 0.571 859 678 500 644 454 4;
  • 79) 0.571 859 678 500 644 454 4 × 2 = 1 + 0.143 719 357 001 288 908 8;
  • 80) 0.143 719 357 001 288 908 8 × 2 = 0 + 0.287 438 714 002 577 817 6;
  • 81) 0.287 438 714 002 577 817 6 × 2 = 0 + 0.574 877 428 005 155 635 2;
  • 82) 0.574 877 428 005 155 635 2 × 2 = 1 + 0.149 754 856 010 311 270 4;
  • 83) 0.149 754 856 010 311 270 4 × 2 = 0 + 0.299 509 712 020 622 540 8;
  • 84) 0.299 509 712 020 622 540 8 × 2 = 0 + 0.599 019 424 041 245 081 6;
  • 85) 0.599 019 424 041 245 081 6 × 2 = 1 + 0.198 038 848 082 490 163 2;
  • 86) 0.198 038 848 082 490 163 2 × 2 = 0 + 0.396 077 696 164 980 326 4;
  • 87) 0.396 077 696 164 980 326 4 × 2 = 0 + 0.792 155 392 329 960 652 8;
  • 88) 0.792 155 392 329 960 652 8 × 2 = 1 + 0.584 310 784 659 921 305 6;
  • 89) 0.584 310 784 659 921 305 6 × 2 = 1 + 0.168 621 569 319 842 611 2;
  • 90) 0.168 621 569 319 842 611 2 × 2 = 0 + 0.337 243 138 639 685 222 4;
  • 91) 0.337 243 138 639 685 222 4 × 2 = 0 + 0.674 486 277 279 370 444 8;
  • 92) 0.674 486 277 279 370 444 8 × 2 = 1 + 0.348 972 554 558 740 889 6;
  • 93) 0.348 972 554 558 740 889 6 × 2 = 0 + 0.697 945 109 117 481 779 2;
  • 94) 0.697 945 109 117 481 779 2 × 2 = 1 + 0.395 890 218 234 963 558 4;
  • 95) 0.395 890 218 234 963 558 4 × 2 = 0 + 0.791 780 436 469 927 116 8;
  • 96) 0.791 780 436 469 927 116 8 × 2 = 1 + 0.583 560 872 939 854 233 6;
  • 97) 0.583 560 872 939 854 233 6 × 2 = 1 + 0.167 121 745 879 708 467 2;
  • 98) 0.167 121 745 879 708 467 2 × 2 = 0 + 0.334 243 491 759 416 934 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:


0.000 000 000 000 014 387 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1011 1100 0110 0000 0000 0010 0100 1001 1001 0101 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 387 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1011 1100 0110 0000 0000 0010 0100 1001 1001 0101 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:


0.000 000 000 000 014 387 6(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1011 1100 0110 0000 0000 0010 0100 1001 1001 0101 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1100 1011 1100 0110 0000 0000 0010 0100 1001 1001 0101 10(2) × 20 =

1.0000 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 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 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 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 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 0110 =

0000 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 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 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 0110


Decimal number -0.000 000 000 000 014 387 6 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0011 0010 1111 0001 1000 0000 0000 1001 0010 0110 0101 0110

» Convert decimal -0.000 000 000 000 014 392 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