-0.000 000 000 000 014 367 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 367 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
-0.000 000 000 000 014 367 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:

|-0.000 000 000 000 014 367 9| = 0.000 000 000 000 014 367 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: 0.000 000 000 000 014 367 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) 0.000 000 000 000 014 367 9 × 2 = 0 + 0.000 000 000 000 028 735 8;
  • 2) 0.000 000 000 000 028 735 8 × 2 = 0 + 0.000 000 000 000 057 471 6;
  • 3) 0.000 000 000 000 057 471 6 × 2 = 0 + 0.000 000 000 000 114 943 2;
  • 4) 0.000 000 000 000 114 943 2 × 2 = 0 + 0.000 000 000 000 229 886 4;
  • 5) 0.000 000 000 000 229 886 4 × 2 = 0 + 0.000 000 000 000 459 772 8;
  • 6) 0.000 000 000 000 459 772 8 × 2 = 0 + 0.000 000 000 000 919 545 6;
  • 7) 0.000 000 000 000 919 545 6 × 2 = 0 + 0.000 000 000 001 839 091 2;
  • 8) 0.000 000 000 001 839 091 2 × 2 = 0 + 0.000 000 000 003 678 182 4;
  • 9) 0.000 000 000 003 678 182 4 × 2 = 0 + 0.000 000 000 007 356 364 8;
  • 10) 0.000 000 000 007 356 364 8 × 2 = 0 + 0.000 000 000 014 712 729 6;
  • 11) 0.000 000 000 014 712 729 6 × 2 = 0 + 0.000 000 000 029 425 459 2;
  • 12) 0.000 000 000 029 425 459 2 × 2 = 0 + 0.000 000 000 058 850 918 4;
  • 13) 0.000 000 000 058 850 918 4 × 2 = 0 + 0.000 000 000 117 701 836 8;
  • 14) 0.000 000 000 117 701 836 8 × 2 = 0 + 0.000 000 000 235 403 673 6;
  • 15) 0.000 000 000 235 403 673 6 × 2 = 0 + 0.000 000 000 470 807 347 2;
  • 16) 0.000 000 000 470 807 347 2 × 2 = 0 + 0.000 000 000 941 614 694 4;
  • 17) 0.000 000 000 941 614 694 4 × 2 = 0 + 0.000 000 001 883 229 388 8;
  • 18) 0.000 000 001 883 229 388 8 × 2 = 0 + 0.000 000 003 766 458 777 6;
  • 19) 0.000 000 003 766 458 777 6 × 2 = 0 + 0.000 000 007 532 917 555 2;
  • 20) 0.000 000 007 532 917 555 2 × 2 = 0 + 0.000 000 015 065 835 110 4;
  • 21) 0.000 000 015 065 835 110 4 × 2 = 0 + 0.000 000 030 131 670 220 8;
  • 22) 0.000 000 030 131 670 220 8 × 2 = 0 + 0.000 000 060 263 340 441 6;
  • 23) 0.000 000 060 263 340 441 6 × 2 = 0 + 0.000 000 120 526 680 883 2;
  • 24) 0.000 000 120 526 680 883 2 × 2 = 0 + 0.000 000 241 053 361 766 4;
  • 25) 0.000 000 241 053 361 766 4 × 2 = 0 + 0.000 000 482 106 723 532 8;
  • 26) 0.000 000 482 106 723 532 8 × 2 = 0 + 0.000 000 964 213 447 065 6;
  • 27) 0.000 000 964 213 447 065 6 × 2 = 0 + 0.000 001 928 426 894 131 2;
  • 28) 0.000 001 928 426 894 131 2 × 2 = 0 + 0.000 003 856 853 788 262 4;
  • 29) 0.000 003 856 853 788 262 4 × 2 = 0 + 0.000 007 713 707 576 524 8;
  • 30) 0.000 007 713 707 576 524 8 × 2 = 0 + 0.000 015 427 415 153 049 6;
  • 31) 0.000 015 427 415 153 049 6 × 2 = 0 + 0.000 030 854 830 306 099 2;
  • 32) 0.000 030 854 830 306 099 2 × 2 = 0 + 0.000 061 709 660 612 198 4;
  • 33) 0.000 061 709 660 612 198 4 × 2 = 0 + 0.000 123 419 321 224 396 8;
  • 34) 0.000 123 419 321 224 396 8 × 2 = 0 + 0.000 246 838 642 448 793 6;
  • 35) 0.000 246 838 642 448 793 6 × 2 = 0 + 0.000 493 677 284 897 587 2;
  • 36) 0.000 493 677 284 897 587 2 × 2 = 0 + 0.000 987 354 569 795 174 4;
  • 37) 0.000 987 354 569 795 174 4 × 2 = 0 + 0.001 974 709 139 590 348 8;
  • 38) 0.001 974 709 139 590 348 8 × 2 = 0 + 0.003 949 418 279 180 697 6;
  • 39) 0.003 949 418 279 180 697 6 × 2 = 0 + 0.007 898 836 558 361 395 2;
  • 40) 0.007 898 836 558 361 395 2 × 2 = 0 + 0.015 797 673 116 722 790 4;
  • 41) 0.015 797 673 116 722 790 4 × 2 = 0 + 0.031 595 346 233 445 580 8;
  • 42) 0.031 595 346 233 445 580 8 × 2 = 0 + 0.063 190 692 466 891 161 6;
  • 43) 0.063 190 692 466 891 161 6 × 2 = 0 + 0.126 381 384 933 782 323 2;
  • 44) 0.126 381 384 933 782 323 2 × 2 = 0 + 0.252 762 769 867 564 646 4;
  • 45) 0.252 762 769 867 564 646 4 × 2 = 0 + 0.505 525 539 735 129 292 8;
  • 46) 0.505 525 539 735 129 292 8 × 2 = 1 + 0.011 051 079 470 258 585 6;
  • 47) 0.011 051 079 470 258 585 6 × 2 = 0 + 0.022 102 158 940 517 171 2;
  • 48) 0.022 102 158 940 517 171 2 × 2 = 0 + 0.044 204 317 881 034 342 4;
  • 49) 0.044 204 317 881 034 342 4 × 2 = 0 + 0.088 408 635 762 068 684 8;
  • 50) 0.088 408 635 762 068 684 8 × 2 = 0 + 0.176 817 271 524 137 369 6;
  • 51) 0.176 817 271 524 137 369 6 × 2 = 0 + 0.353 634 543 048 274 739 2;
  • 52) 0.353 634 543 048 274 739 2 × 2 = 0 + 0.707 269 086 096 549 478 4;
  • 53) 0.707 269 086 096 549 478 4 × 2 = 1 + 0.414 538 172 193 098 956 8;
  • 54) 0.414 538 172 193 098 956 8 × 2 = 0 + 0.829 076 344 386 197 913 6;
  • 55) 0.829 076 344 386 197 913 6 × 2 = 1 + 0.658 152 688 772 395 827 2;
  • 56) 0.658 152 688 772 395 827 2 × 2 = 1 + 0.316 305 377 544 791 654 4;
  • 57) 0.316 305 377 544 791 654 4 × 2 = 0 + 0.632 610 755 089 583 308 8;
  • 58) 0.632 610 755 089 583 308 8 × 2 = 1 + 0.265 221 510 179 166 617 6;
  • 59) 0.265 221 510 179 166 617 6 × 2 = 0 + 0.530 443 020 358 333 235 2;
  • 60) 0.530 443 020 358 333 235 2 × 2 = 1 + 0.060 886 040 716 666 470 4;
  • 61) 0.060 886 040 716 666 470 4 × 2 = 0 + 0.121 772 081 433 332 940 8;
  • 62) 0.121 772 081 433 332 940 8 × 2 = 0 + 0.243 544 162 866 665 881 6;
  • 63) 0.243 544 162 866 665 881 6 × 2 = 0 + 0.487 088 325 733 331 763 2;
  • 64) 0.487 088 325 733 331 763 2 × 2 = 0 + 0.974 176 651 466 663 526 4;
  • 65) 0.974 176 651 466 663 526 4 × 2 = 1 + 0.948 353 302 933 327 052 8;
  • 66) 0.948 353 302 933 327 052 8 × 2 = 1 + 0.896 706 605 866 654 105 6;
  • 67) 0.896 706 605 866 654 105 6 × 2 = 1 + 0.793 413 211 733 308 211 2;
  • 68) 0.793 413 211 733 308 211 2 × 2 = 1 + 0.586 826 423 466 616 422 4;
  • 69) 0.586 826 423 466 616 422 4 × 2 = 1 + 0.173 652 846 933 232 844 8;
  • 70) 0.173 652 846 933 232 844 8 × 2 = 0 + 0.347 305 693 866 465 689 6;
  • 71) 0.347 305 693 866 465 689 6 × 2 = 0 + 0.694 611 387 732 931 379 2;
  • 72) 0.694 611 387 732 931 379 2 × 2 = 1 + 0.389 222 775 465 862 758 4;
  • 73) 0.389 222 775 465 862 758 4 × 2 = 0 + 0.778 445 550 931 725 516 8;
  • 74) 0.778 445 550 931 725 516 8 × 2 = 1 + 0.556 891 101 863 451 033 6;
  • 75) 0.556 891 101 863 451 033 6 × 2 = 1 + 0.113 782 203 726 902 067 2;
  • 76) 0.113 782 203 726 902 067 2 × 2 = 0 + 0.227 564 407 453 804 134 4;
  • 77) 0.227 564 407 453 804 134 4 × 2 = 0 + 0.455 128 814 907 608 268 8;
  • 78) 0.455 128 814 907 608 268 8 × 2 = 0 + 0.910 257 629 815 216 537 6;
  • 79) 0.910 257 629 815 216 537 6 × 2 = 1 + 0.820 515 259 630 433 075 2;
  • 80) 0.820 515 259 630 433 075 2 × 2 = 1 + 0.641 030 519 260 866 150 4;
  • 81) 0.641 030 519 260 866 150 4 × 2 = 1 + 0.282 061 038 521 732 300 8;
  • 82) 0.282 061 038 521 732 300 8 × 2 = 0 + 0.564 122 077 043 464 601 6;
  • 83) 0.564 122 077 043 464 601 6 × 2 = 1 + 0.128 244 154 086 929 203 2;
  • 84) 0.128 244 154 086 929 203 2 × 2 = 0 + 0.256 488 308 173 858 406 4;
  • 85) 0.256 488 308 173 858 406 4 × 2 = 0 + 0.512 976 616 347 716 812 8;
  • 86) 0.512 976 616 347 716 812 8 × 2 = 1 + 0.025 953 232 695 433 625 6;
  • 87) 0.025 953 232 695 433 625 6 × 2 = 0 + 0.051 906 465 390 867 251 2;
  • 88) 0.051 906 465 390 867 251 2 × 2 = 0 + 0.103 812 930 781 734 502 4;
  • 89) 0.103 812 930 781 734 502 4 × 2 = 0 + 0.207 625 861 563 469 004 8;
  • 90) 0.207 625 861 563 469 004 8 × 2 = 0 + 0.415 251 723 126 938 009 6;
  • 91) 0.415 251 723 126 938 009 6 × 2 = 0 + 0.830 503 446 253 876 019 2;
  • 92) 0.830 503 446 253 876 019 2 × 2 = 1 + 0.661 006 892 507 752 038 4;
  • 93) 0.661 006 892 507 752 038 4 × 2 = 1 + 0.322 013 785 015 504 076 8;
  • 94) 0.322 013 785 015 504 076 8 × 2 = 0 + 0.644 027 570 031 008 153 6;
  • 95) 0.644 027 570 031 008 153 6 × 2 = 1 + 0.288 055 140 062 016 307 2;
  • 96) 0.288 055 140 062 016 307 2 × 2 = 0 + 0.576 110 280 124 032 614 4;
  • 97) 0.576 110 280 124 032 614 4 × 2 = 1 + 0.152 220 560 248 065 228 8;
  • 98) 0.152 220 560 248 065 228 8 × 2 = 0 + 0.304 441 120 496 130 457 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:


0.000 000 000 000 014 367 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0101 0000 1111 1001 0110 0011 1010 0100 0001 1010 10(2)

6. Positive number before normalization:

0.000 000 000 000 014 367 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0101 0000 1111 1001 0110 0011 1010 0100 0001 1010 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:


0.000 000 000 000 014 367 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0101 0000 1111 1001 0110 0011 1010 0100 0001 1010 10(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0101 0000 1111 1001 0110 0011 1010 0100 0001 1010 10(2) × 20 =

1.0000 0010 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010(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 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010


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 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010 =

0000 0010 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010


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 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010


Decimal number -0.000 000 000 000 014 367 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1101 0100 0011 1110 0101 1000 1110 1001 0000 0110 1010

» Convert decimal -0.000 000 000 000 014 373 8 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