-0.000 000 000 000 014 364 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 014 364 2(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
-0.000 000 000 000 014 364 2(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:

|-0.000 000 000 000 014 364 2| = 0.000 000 000 000 014 364 2


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: 0.000 000 000 000 014 364 2.

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) 0.000 000 000 000 014 364 2 × 2 = 0 + 0.000 000 000 000 028 728 4;
  • 2) 0.000 000 000 000 028 728 4 × 2 = 0 + 0.000 000 000 000 057 456 8;
  • 3) 0.000 000 000 000 057 456 8 × 2 = 0 + 0.000 000 000 000 114 913 6;
  • 4) 0.000 000 000 000 114 913 6 × 2 = 0 + 0.000 000 000 000 229 827 2;
  • 5) 0.000 000 000 000 229 827 2 × 2 = 0 + 0.000 000 000 000 459 654 4;
  • 6) 0.000 000 000 000 459 654 4 × 2 = 0 + 0.000 000 000 000 919 308 8;
  • 7) 0.000 000 000 000 919 308 8 × 2 = 0 + 0.000 000 000 001 838 617 6;
  • 8) 0.000 000 000 001 838 617 6 × 2 = 0 + 0.000 000 000 003 677 235 2;
  • 9) 0.000 000 000 003 677 235 2 × 2 = 0 + 0.000 000 000 007 354 470 4;
  • 10) 0.000 000 000 007 354 470 4 × 2 = 0 + 0.000 000 000 014 708 940 8;
  • 11) 0.000 000 000 014 708 940 8 × 2 = 0 + 0.000 000 000 029 417 881 6;
  • 12) 0.000 000 000 029 417 881 6 × 2 = 0 + 0.000 000 000 058 835 763 2;
  • 13) 0.000 000 000 058 835 763 2 × 2 = 0 + 0.000 000 000 117 671 526 4;
  • 14) 0.000 000 000 117 671 526 4 × 2 = 0 + 0.000 000 000 235 343 052 8;
  • 15) 0.000 000 000 235 343 052 8 × 2 = 0 + 0.000 000 000 470 686 105 6;
  • 16) 0.000 000 000 470 686 105 6 × 2 = 0 + 0.000 000 000 941 372 211 2;
  • 17) 0.000 000 000 941 372 211 2 × 2 = 0 + 0.000 000 001 882 744 422 4;
  • 18) 0.000 000 001 882 744 422 4 × 2 = 0 + 0.000 000 003 765 488 844 8;
  • 19) 0.000 000 003 765 488 844 8 × 2 = 0 + 0.000 000 007 530 977 689 6;
  • 20) 0.000 000 007 530 977 689 6 × 2 = 0 + 0.000 000 015 061 955 379 2;
  • 21) 0.000 000 015 061 955 379 2 × 2 = 0 + 0.000 000 030 123 910 758 4;
  • 22) 0.000 000 030 123 910 758 4 × 2 = 0 + 0.000 000 060 247 821 516 8;
  • 23) 0.000 000 060 247 821 516 8 × 2 = 0 + 0.000 000 120 495 643 033 6;
  • 24) 0.000 000 120 495 643 033 6 × 2 = 0 + 0.000 000 240 991 286 067 2;
  • 25) 0.000 000 240 991 286 067 2 × 2 = 0 + 0.000 000 481 982 572 134 4;
  • 26) 0.000 000 481 982 572 134 4 × 2 = 0 + 0.000 000 963 965 144 268 8;
  • 27) 0.000 000 963 965 144 268 8 × 2 = 0 + 0.000 001 927 930 288 537 6;
  • 28) 0.000 001 927 930 288 537 6 × 2 = 0 + 0.000 003 855 860 577 075 2;
  • 29) 0.000 003 855 860 577 075 2 × 2 = 0 + 0.000 007 711 721 154 150 4;
  • 30) 0.000 007 711 721 154 150 4 × 2 = 0 + 0.000 015 423 442 308 300 8;
  • 31) 0.000 015 423 442 308 300 8 × 2 = 0 + 0.000 030 846 884 616 601 6;
  • 32) 0.000 030 846 884 616 601 6 × 2 = 0 + 0.000 061 693 769 233 203 2;
  • 33) 0.000 061 693 769 233 203 2 × 2 = 0 + 0.000 123 387 538 466 406 4;
  • 34) 0.000 123 387 538 466 406 4 × 2 = 0 + 0.000 246 775 076 932 812 8;
  • 35) 0.000 246 775 076 932 812 8 × 2 = 0 + 0.000 493 550 153 865 625 6;
  • 36) 0.000 493 550 153 865 625 6 × 2 = 0 + 0.000 987 100 307 731 251 2;
  • 37) 0.000 987 100 307 731 251 2 × 2 = 0 + 0.001 974 200 615 462 502 4;
  • 38) 0.001 974 200 615 462 502 4 × 2 = 0 + 0.003 948 401 230 925 004 8;
  • 39) 0.003 948 401 230 925 004 8 × 2 = 0 + 0.007 896 802 461 850 009 6;
  • 40) 0.007 896 802 461 850 009 6 × 2 = 0 + 0.015 793 604 923 700 019 2;
  • 41) 0.015 793 604 923 700 019 2 × 2 = 0 + 0.031 587 209 847 400 038 4;
  • 42) 0.031 587 209 847 400 038 4 × 2 = 0 + 0.063 174 419 694 800 076 8;
  • 43) 0.063 174 419 694 800 076 8 × 2 = 0 + 0.126 348 839 389 600 153 6;
  • 44) 0.126 348 839 389 600 153 6 × 2 = 0 + 0.252 697 678 779 200 307 2;
  • 45) 0.252 697 678 779 200 307 2 × 2 = 0 + 0.505 395 357 558 400 614 4;
  • 46) 0.505 395 357 558 400 614 4 × 2 = 1 + 0.010 790 715 116 801 228 8;
  • 47) 0.010 790 715 116 801 228 8 × 2 = 0 + 0.021 581 430 233 602 457 6;
  • 48) 0.021 581 430 233 602 457 6 × 2 = 0 + 0.043 162 860 467 204 915 2;
  • 49) 0.043 162 860 467 204 915 2 × 2 = 0 + 0.086 325 720 934 409 830 4;
  • 50) 0.086 325 720 934 409 830 4 × 2 = 0 + 0.172 651 441 868 819 660 8;
  • 51) 0.172 651 441 868 819 660 8 × 2 = 0 + 0.345 302 883 737 639 321 6;
  • 52) 0.345 302 883 737 639 321 6 × 2 = 0 + 0.690 605 767 475 278 643 2;
  • 53) 0.690 605 767 475 278 643 2 × 2 = 1 + 0.381 211 534 950 557 286 4;
  • 54) 0.381 211 534 950 557 286 4 × 2 = 0 + 0.762 423 069 901 114 572 8;
  • 55) 0.762 423 069 901 114 572 8 × 2 = 1 + 0.524 846 139 802 229 145 6;
  • 56) 0.524 846 139 802 229 145 6 × 2 = 1 + 0.049 692 279 604 458 291 2;
  • 57) 0.049 692 279 604 458 291 2 × 2 = 0 + 0.099 384 559 208 916 582 4;
  • 58) 0.099 384 559 208 916 582 4 × 2 = 0 + 0.198 769 118 417 833 164 8;
  • 59) 0.198 769 118 417 833 164 8 × 2 = 0 + 0.397 538 236 835 666 329 6;
  • 60) 0.397 538 236 835 666 329 6 × 2 = 0 + 0.795 076 473 671 332 659 2;
  • 61) 0.795 076 473 671 332 659 2 × 2 = 1 + 0.590 152 947 342 665 318 4;
  • 62) 0.590 152 947 342 665 318 4 × 2 = 1 + 0.180 305 894 685 330 636 8;
  • 63) 0.180 305 894 685 330 636 8 × 2 = 0 + 0.360 611 789 370 661 273 6;
  • 64) 0.360 611 789 370 661 273 6 × 2 = 0 + 0.721 223 578 741 322 547 2;
  • 65) 0.721 223 578 741 322 547 2 × 2 = 1 + 0.442 447 157 482 645 094 4;
  • 66) 0.442 447 157 482 645 094 4 × 2 = 0 + 0.884 894 314 965 290 188 8;
  • 67) 0.884 894 314 965 290 188 8 × 2 = 1 + 0.769 788 629 930 580 377 6;
  • 68) 0.769 788 629 930 580 377 6 × 2 = 1 + 0.539 577 259 861 160 755 2;
  • 69) 0.539 577 259 861 160 755 2 × 2 = 1 + 0.079 154 519 722 321 510 4;
  • 70) 0.079 154 519 722 321 510 4 × 2 = 0 + 0.158 309 039 444 643 020 8;
  • 71) 0.158 309 039 444 643 020 8 × 2 = 0 + 0.316 618 078 889 286 041 6;
  • 72) 0.316 618 078 889 286 041 6 × 2 = 0 + 0.633 236 157 778 572 083 2;
  • 73) 0.633 236 157 778 572 083 2 × 2 = 1 + 0.266 472 315 557 144 166 4;
  • 74) 0.266 472 315 557 144 166 4 × 2 = 0 + 0.532 944 631 114 288 332 8;
  • 75) 0.532 944 631 114 288 332 8 × 2 = 1 + 0.065 889 262 228 576 665 6;
  • 76) 0.065 889 262 228 576 665 6 × 2 = 0 + 0.131 778 524 457 153 331 2;
  • 77) 0.131 778 524 457 153 331 2 × 2 = 0 + 0.263 557 048 914 306 662 4;
  • 78) 0.263 557 048 914 306 662 4 × 2 = 0 + 0.527 114 097 828 613 324 8;
  • 79) 0.527 114 097 828 613 324 8 × 2 = 1 + 0.054 228 195 657 226 649 6;
  • 80) 0.054 228 195 657 226 649 6 × 2 = 0 + 0.108 456 391 314 453 299 2;
  • 81) 0.108 456 391 314 453 299 2 × 2 = 0 + 0.216 912 782 628 906 598 4;
  • 82) 0.216 912 782 628 906 598 4 × 2 = 0 + 0.433 825 565 257 813 196 8;
  • 83) 0.433 825 565 257 813 196 8 × 2 = 0 + 0.867 651 130 515 626 393 6;
  • 84) 0.867 651 130 515 626 393 6 × 2 = 1 + 0.735 302 261 031 252 787 2;
  • 85) 0.735 302 261 031 252 787 2 × 2 = 1 + 0.470 604 522 062 505 574 4;
  • 86) 0.470 604 522 062 505 574 4 × 2 = 0 + 0.941 209 044 125 011 148 8;
  • 87) 0.941 209 044 125 011 148 8 × 2 = 1 + 0.882 418 088 250 022 297 6;
  • 88) 0.882 418 088 250 022 297 6 × 2 = 1 + 0.764 836 176 500 044 595 2;
  • 89) 0.764 836 176 500 044 595 2 × 2 = 1 + 0.529 672 353 000 089 190 4;
  • 90) 0.529 672 353 000 089 190 4 × 2 = 1 + 0.059 344 706 000 178 380 8;
  • 91) 0.059 344 706 000 178 380 8 × 2 = 0 + 0.118 689 412 000 356 761 6;
  • 92) 0.118 689 412 000 356 761 6 × 2 = 0 + 0.237 378 824 000 713 523 2;
  • 93) 0.237 378 824 000 713 523 2 × 2 = 0 + 0.474 757 648 001 427 046 4;
  • 94) 0.474 757 648 001 427 046 4 × 2 = 0 + 0.949 515 296 002 854 092 8;
  • 95) 0.949 515 296 002 854 092 8 × 2 = 1 + 0.899 030 592 005 708 185 6;
  • 96) 0.899 030 592 005 708 185 6 × 2 = 1 + 0.798 061 184 011 416 371 2;
  • 97) 0.798 061 184 011 416 371 2 × 2 = 1 + 0.596 122 368 022 832 742 4;
  • 98) 0.596 122 368 022 832 742 4 × 2 = 1 + 0.192 244 736 045 665 484 8;

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:


0.000 000 000 000 014 364 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0000 1100 1011 1000 1010 0010 0001 1011 1100 0011 11(2)

6. Positive number before normalization:

0.000 000 000 000 014 364 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0000 1100 1011 1000 1010 0010 0001 1011 1100 0011 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:


0.000 000 000 000 014 364 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0000 1100 1011 1000 1010 0010 0001 1011 1100 0011 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 0000 1011 0000 1100 1011 1000 1010 0010 0001 1011 1100 0011 11(2) × 20 =

1.0000 0010 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111(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 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111


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 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111 =

0000 0010 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111


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 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111


Decimal number -0.000 000 000 000 014 364 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0001 - 0000 0010 1100 0011 0010 1110 0010 1000 1000 0110 1111 0000 1111

» Convert decimal -0.000 000 000 000 014 364 6 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