-0.000 000 000 000 176 557 496 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 496(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
-0.000 000 000 000 176 557 496(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:

|-0.000 000 000 000 176 557 496| = 0.000 000 000 000 176 557 496


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: 0.000 000 000 000 176 557 496.

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) 0.000 000 000 000 176 557 496 × 2 = 0 + 0.000 000 000 000 353 114 992;
  • 2) 0.000 000 000 000 353 114 992 × 2 = 0 + 0.000 000 000 000 706 229 984;
  • 3) 0.000 000 000 000 706 229 984 × 2 = 0 + 0.000 000 000 001 412 459 968;
  • 4) 0.000 000 000 001 412 459 968 × 2 = 0 + 0.000 000 000 002 824 919 936;
  • 5) 0.000 000 000 002 824 919 936 × 2 = 0 + 0.000 000 000 005 649 839 872;
  • 6) 0.000 000 000 005 649 839 872 × 2 = 0 + 0.000 000 000 011 299 679 744;
  • 7) 0.000 000 000 011 299 679 744 × 2 = 0 + 0.000 000 000 022 599 359 488;
  • 8) 0.000 000 000 022 599 359 488 × 2 = 0 + 0.000 000 000 045 198 718 976;
  • 9) 0.000 000 000 045 198 718 976 × 2 = 0 + 0.000 000 000 090 397 437 952;
  • 10) 0.000 000 000 090 397 437 952 × 2 = 0 + 0.000 000 000 180 794 875 904;
  • 11) 0.000 000 000 180 794 875 904 × 2 = 0 + 0.000 000 000 361 589 751 808;
  • 12) 0.000 000 000 361 589 751 808 × 2 = 0 + 0.000 000 000 723 179 503 616;
  • 13) 0.000 000 000 723 179 503 616 × 2 = 0 + 0.000 000 001 446 359 007 232;
  • 14) 0.000 000 001 446 359 007 232 × 2 = 0 + 0.000 000 002 892 718 014 464;
  • 15) 0.000 000 002 892 718 014 464 × 2 = 0 + 0.000 000 005 785 436 028 928;
  • 16) 0.000 000 005 785 436 028 928 × 2 = 0 + 0.000 000 011 570 872 057 856;
  • 17) 0.000 000 011 570 872 057 856 × 2 = 0 + 0.000 000 023 141 744 115 712;
  • 18) 0.000 000 023 141 744 115 712 × 2 = 0 + 0.000 000 046 283 488 231 424;
  • 19) 0.000 000 046 283 488 231 424 × 2 = 0 + 0.000 000 092 566 976 462 848;
  • 20) 0.000 000 092 566 976 462 848 × 2 = 0 + 0.000 000 185 133 952 925 696;
  • 21) 0.000 000 185 133 952 925 696 × 2 = 0 + 0.000 000 370 267 905 851 392;
  • 22) 0.000 000 370 267 905 851 392 × 2 = 0 + 0.000 000 740 535 811 702 784;
  • 23) 0.000 000 740 535 811 702 784 × 2 = 0 + 0.000 001 481 071 623 405 568;
  • 24) 0.000 001 481 071 623 405 568 × 2 = 0 + 0.000 002 962 143 246 811 136;
  • 25) 0.000 002 962 143 246 811 136 × 2 = 0 + 0.000 005 924 286 493 622 272;
  • 26) 0.000 005 924 286 493 622 272 × 2 = 0 + 0.000 011 848 572 987 244 544;
  • 27) 0.000 011 848 572 987 244 544 × 2 = 0 + 0.000 023 697 145 974 489 088;
  • 28) 0.000 023 697 145 974 489 088 × 2 = 0 + 0.000 047 394 291 948 978 176;
  • 29) 0.000 047 394 291 948 978 176 × 2 = 0 + 0.000 094 788 583 897 956 352;
  • 30) 0.000 094 788 583 897 956 352 × 2 = 0 + 0.000 189 577 167 795 912 704;
  • 31) 0.000 189 577 167 795 912 704 × 2 = 0 + 0.000 379 154 335 591 825 408;
  • 32) 0.000 379 154 335 591 825 408 × 2 = 0 + 0.000 758 308 671 183 650 816;
  • 33) 0.000 758 308 671 183 650 816 × 2 = 0 + 0.001 516 617 342 367 301 632;
  • 34) 0.001 516 617 342 367 301 632 × 2 = 0 + 0.003 033 234 684 734 603 264;
  • 35) 0.003 033 234 684 734 603 264 × 2 = 0 + 0.006 066 469 369 469 206 528;
  • 36) 0.006 066 469 369 469 206 528 × 2 = 0 + 0.012 132 938 738 938 413 056;
  • 37) 0.012 132 938 738 938 413 056 × 2 = 0 + 0.024 265 877 477 876 826 112;
  • 38) 0.024 265 877 477 876 826 112 × 2 = 0 + 0.048 531 754 955 753 652 224;
  • 39) 0.048 531 754 955 753 652 224 × 2 = 0 + 0.097 063 509 911 507 304 448;
  • 40) 0.097 063 509 911 507 304 448 × 2 = 0 + 0.194 127 019 823 014 608 896;
  • 41) 0.194 127 019 823 014 608 896 × 2 = 0 + 0.388 254 039 646 029 217 792;
  • 42) 0.388 254 039 646 029 217 792 × 2 = 0 + 0.776 508 079 292 058 435 584;
  • 43) 0.776 508 079 292 058 435 584 × 2 = 1 + 0.553 016 158 584 116 871 168;
  • 44) 0.553 016 158 584 116 871 168 × 2 = 1 + 0.106 032 317 168 233 742 336;
  • 45) 0.106 032 317 168 233 742 336 × 2 = 0 + 0.212 064 634 336 467 484 672;
  • 46) 0.212 064 634 336 467 484 672 × 2 = 0 + 0.424 129 268 672 934 969 344;
  • 47) 0.424 129 268 672 934 969 344 × 2 = 0 + 0.848 258 537 345 869 938 688;
  • 48) 0.848 258 537 345 869 938 688 × 2 = 1 + 0.696 517 074 691 739 877 376;
  • 49) 0.696 517 074 691 739 877 376 × 2 = 1 + 0.393 034 149 383 479 754 752;
  • 50) 0.393 034 149 383 479 754 752 × 2 = 0 + 0.786 068 298 766 959 509 504;
  • 51) 0.786 068 298 766 959 509 504 × 2 = 1 + 0.572 136 597 533 919 019 008;
  • 52) 0.572 136 597 533 919 019 008 × 2 = 1 + 0.144 273 195 067 838 038 016;
  • 53) 0.144 273 195 067 838 038 016 × 2 = 0 + 0.288 546 390 135 676 076 032;
  • 54) 0.288 546 390 135 676 076 032 × 2 = 0 + 0.577 092 780 271 352 152 064;
  • 55) 0.577 092 780 271 352 152 064 × 2 = 1 + 0.154 185 560 542 704 304 128;
  • 56) 0.154 185 560 542 704 304 128 × 2 = 0 + 0.308 371 121 085 408 608 256;
  • 57) 0.308 371 121 085 408 608 256 × 2 = 0 + 0.616 742 242 170 817 216 512;
  • 58) 0.616 742 242 170 817 216 512 × 2 = 1 + 0.233 484 484 341 634 433 024;
  • 59) 0.233 484 484 341 634 433 024 × 2 = 0 + 0.466 968 968 683 268 866 048;
  • 60) 0.466 968 968 683 268 866 048 × 2 = 0 + 0.933 937 937 366 537 732 096;
  • 61) 0.933 937 937 366 537 732 096 × 2 = 1 + 0.867 875 874 733 075 464 192;
  • 62) 0.867 875 874 733 075 464 192 × 2 = 1 + 0.735 751 749 466 150 928 384;
  • 63) 0.735 751 749 466 150 928 384 × 2 = 1 + 0.471 503 498 932 301 856 768;
  • 64) 0.471 503 498 932 301 856 768 × 2 = 0 + 0.943 006 997 864 603 713 536;
  • 65) 0.943 006 997 864 603 713 536 × 2 = 1 + 0.886 013 995 729 207 427 072;
  • 66) 0.886 013 995 729 207 427 072 × 2 = 1 + 0.772 027 991 458 414 854 144;
  • 67) 0.772 027 991 458 414 854 144 × 2 = 1 + 0.544 055 982 916 829 708 288;
  • 68) 0.544 055 982 916 829 708 288 × 2 = 1 + 0.088 111 965 833 659 416 576;
  • 69) 0.088 111 965 833 659 416 576 × 2 = 0 + 0.176 223 931 667 318 833 152;
  • 70) 0.176 223 931 667 318 833 152 × 2 = 0 + 0.352 447 863 334 637 666 304;
  • 71) 0.352 447 863 334 637 666 304 × 2 = 0 + 0.704 895 726 669 275 332 608;
  • 72) 0.704 895 726 669 275 332 608 × 2 = 1 + 0.409 791 453 338 550 665 216;
  • 73) 0.409 791 453 338 550 665 216 × 2 = 0 + 0.819 582 906 677 101 330 432;
  • 74) 0.819 582 906 677 101 330 432 × 2 = 1 + 0.639 165 813 354 202 660 864;
  • 75) 0.639 165 813 354 202 660 864 × 2 = 1 + 0.278 331 626 708 405 321 728;
  • 76) 0.278 331 626 708 405 321 728 × 2 = 0 + 0.556 663 253 416 810 643 456;
  • 77) 0.556 663 253 416 810 643 456 × 2 = 1 + 0.113 326 506 833 621 286 912;
  • 78) 0.113 326 506 833 621 286 912 × 2 = 0 + 0.226 653 013 667 242 573 824;
  • 79) 0.226 653 013 667 242 573 824 × 2 = 0 + 0.453 306 027 334 485 147 648;
  • 80) 0.453 306 027 334 485 147 648 × 2 = 0 + 0.906 612 054 668 970 295 296;
  • 81) 0.906 612 054 668 970 295 296 × 2 = 1 + 0.813 224 109 337 940 590 592;
  • 82) 0.813 224 109 337 940 590 592 × 2 = 1 + 0.626 448 218 675 881 181 184;
  • 83) 0.626 448 218 675 881 181 184 × 2 = 1 + 0.252 896 437 351 762 362 368;
  • 84) 0.252 896 437 351 762 362 368 × 2 = 0 + 0.505 792 874 703 524 724 736;
  • 85) 0.505 792 874 703 524 724 736 × 2 = 1 + 0.011 585 749 407 049 449 472;
  • 86) 0.011 585 749 407 049 449 472 × 2 = 0 + 0.023 171 498 814 098 898 944;
  • 87) 0.023 171 498 814 098 898 944 × 2 = 0 + 0.046 342 997 628 197 797 888;
  • 88) 0.046 342 997 628 197 797 888 × 2 = 0 + 0.092 685 995 256 395 595 776;
  • 89) 0.092 685 995 256 395 595 776 × 2 = 0 + 0.185 371 990 512 791 191 552;
  • 90) 0.185 371 990 512 791 191 552 × 2 = 0 + 0.370 743 981 025 582 383 104;
  • 91) 0.370 743 981 025 582 383 104 × 2 = 0 + 0.741 487 962 051 164 766 208;
  • 92) 0.741 487 962 051 164 766 208 × 2 = 1 + 0.482 975 924 102 329 532 416;
  • 93) 0.482 975 924 102 329 532 416 × 2 = 0 + 0.965 951 848 204 659 064 832;
  • 94) 0.965 951 848 204 659 064 832 × 2 = 1 + 0.931 903 696 409 318 129 664;
  • 95) 0.931 903 696 409 318 129 664 × 2 = 1 + 0.863 807 392 818 636 259 328;

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:


0.000 000 000 000 176 557 496(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0001 0110 1000 1110 1000 0001 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 496(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0001 0110 1000 1110 1000 0001 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 43 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 176 557 496(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0001 0110 1000 1110 1000 0001 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1111 0001 0110 1000 1110 1000 0001 011(2) × 20 =

1.1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 1011(2) × 2-43


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): -43

Mantissa (not normalized):
1.1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 1011


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-43 + 2(11-1) - 1 =

(-43 + 1 023)(10) =

980(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;
  • 980 ÷ 2 = 490 + 0;
  • 490 ÷ 2 = 245 + 0;
  • 245 ÷ 2 = 122 + 1;
  • 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) =

980(10) =

011 1101 0100(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. 1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 1011 =

1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 1011


Decimal number -0.000 000 000 000 176 557 496 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0111 1000 1011 0100 0111 0100 0000 1011

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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