-0.000 000 000 000 176 557 321 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 321(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
-0.000 000 000 000 176 557 321(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:

|-0.000 000 000 000 176 557 321| = 0.000 000 000 000 176 557 321


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: 0.000 000 000 000 176 557 321.

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) 0.000 000 000 000 176 557 321 × 2 = 0 + 0.000 000 000 000 353 114 642;
  • 2) 0.000 000 000 000 353 114 642 × 2 = 0 + 0.000 000 000 000 706 229 284;
  • 3) 0.000 000 000 000 706 229 284 × 2 = 0 + 0.000 000 000 001 412 458 568;
  • 4) 0.000 000 000 001 412 458 568 × 2 = 0 + 0.000 000 000 002 824 917 136;
  • 5) 0.000 000 000 002 824 917 136 × 2 = 0 + 0.000 000 000 005 649 834 272;
  • 6) 0.000 000 000 005 649 834 272 × 2 = 0 + 0.000 000 000 011 299 668 544;
  • 7) 0.000 000 000 011 299 668 544 × 2 = 0 + 0.000 000 000 022 599 337 088;
  • 8) 0.000 000 000 022 599 337 088 × 2 = 0 + 0.000 000 000 045 198 674 176;
  • 9) 0.000 000 000 045 198 674 176 × 2 = 0 + 0.000 000 000 090 397 348 352;
  • 10) 0.000 000 000 090 397 348 352 × 2 = 0 + 0.000 000 000 180 794 696 704;
  • 11) 0.000 000 000 180 794 696 704 × 2 = 0 + 0.000 000 000 361 589 393 408;
  • 12) 0.000 000 000 361 589 393 408 × 2 = 0 + 0.000 000 000 723 178 786 816;
  • 13) 0.000 000 000 723 178 786 816 × 2 = 0 + 0.000 000 001 446 357 573 632;
  • 14) 0.000 000 001 446 357 573 632 × 2 = 0 + 0.000 000 002 892 715 147 264;
  • 15) 0.000 000 002 892 715 147 264 × 2 = 0 + 0.000 000 005 785 430 294 528;
  • 16) 0.000 000 005 785 430 294 528 × 2 = 0 + 0.000 000 011 570 860 589 056;
  • 17) 0.000 000 011 570 860 589 056 × 2 = 0 + 0.000 000 023 141 721 178 112;
  • 18) 0.000 000 023 141 721 178 112 × 2 = 0 + 0.000 000 046 283 442 356 224;
  • 19) 0.000 000 046 283 442 356 224 × 2 = 0 + 0.000 000 092 566 884 712 448;
  • 20) 0.000 000 092 566 884 712 448 × 2 = 0 + 0.000 000 185 133 769 424 896;
  • 21) 0.000 000 185 133 769 424 896 × 2 = 0 + 0.000 000 370 267 538 849 792;
  • 22) 0.000 000 370 267 538 849 792 × 2 = 0 + 0.000 000 740 535 077 699 584;
  • 23) 0.000 000 740 535 077 699 584 × 2 = 0 + 0.000 001 481 070 155 399 168;
  • 24) 0.000 001 481 070 155 399 168 × 2 = 0 + 0.000 002 962 140 310 798 336;
  • 25) 0.000 002 962 140 310 798 336 × 2 = 0 + 0.000 005 924 280 621 596 672;
  • 26) 0.000 005 924 280 621 596 672 × 2 = 0 + 0.000 011 848 561 243 193 344;
  • 27) 0.000 011 848 561 243 193 344 × 2 = 0 + 0.000 023 697 122 486 386 688;
  • 28) 0.000 023 697 122 486 386 688 × 2 = 0 + 0.000 047 394 244 972 773 376;
  • 29) 0.000 047 394 244 972 773 376 × 2 = 0 + 0.000 094 788 489 945 546 752;
  • 30) 0.000 094 788 489 945 546 752 × 2 = 0 + 0.000 189 576 979 891 093 504;
  • 31) 0.000 189 576 979 891 093 504 × 2 = 0 + 0.000 379 153 959 782 187 008;
  • 32) 0.000 379 153 959 782 187 008 × 2 = 0 + 0.000 758 307 919 564 374 016;
  • 33) 0.000 758 307 919 564 374 016 × 2 = 0 + 0.001 516 615 839 128 748 032;
  • 34) 0.001 516 615 839 128 748 032 × 2 = 0 + 0.003 033 231 678 257 496 064;
  • 35) 0.003 033 231 678 257 496 064 × 2 = 0 + 0.006 066 463 356 514 992 128;
  • 36) 0.006 066 463 356 514 992 128 × 2 = 0 + 0.012 132 926 713 029 984 256;
  • 37) 0.012 132 926 713 029 984 256 × 2 = 0 + 0.024 265 853 426 059 968 512;
  • 38) 0.024 265 853 426 059 968 512 × 2 = 0 + 0.048 531 706 852 119 937 024;
  • 39) 0.048 531 706 852 119 937 024 × 2 = 0 + 0.097 063 413 704 239 874 048;
  • 40) 0.097 063 413 704 239 874 048 × 2 = 0 + 0.194 126 827 408 479 748 096;
  • 41) 0.194 126 827 408 479 748 096 × 2 = 0 + 0.388 253 654 816 959 496 192;
  • 42) 0.388 253 654 816 959 496 192 × 2 = 0 + 0.776 507 309 633 918 992 384;
  • 43) 0.776 507 309 633 918 992 384 × 2 = 1 + 0.553 014 619 267 837 984 768;
  • 44) 0.553 014 619 267 837 984 768 × 2 = 1 + 0.106 029 238 535 675 969 536;
  • 45) 0.106 029 238 535 675 969 536 × 2 = 0 + 0.212 058 477 071 351 939 072;
  • 46) 0.212 058 477 071 351 939 072 × 2 = 0 + 0.424 116 954 142 703 878 144;
  • 47) 0.424 116 954 142 703 878 144 × 2 = 0 + 0.848 233 908 285 407 756 288;
  • 48) 0.848 233 908 285 407 756 288 × 2 = 1 + 0.696 467 816 570 815 512 576;
  • 49) 0.696 467 816 570 815 512 576 × 2 = 1 + 0.392 935 633 141 631 025 152;
  • 50) 0.392 935 633 141 631 025 152 × 2 = 0 + 0.785 871 266 283 262 050 304;
  • 51) 0.785 871 266 283 262 050 304 × 2 = 1 + 0.571 742 532 566 524 100 608;
  • 52) 0.571 742 532 566 524 100 608 × 2 = 1 + 0.143 485 065 133 048 201 216;
  • 53) 0.143 485 065 133 048 201 216 × 2 = 0 + 0.286 970 130 266 096 402 432;
  • 54) 0.286 970 130 266 096 402 432 × 2 = 0 + 0.573 940 260 532 192 804 864;
  • 55) 0.573 940 260 532 192 804 864 × 2 = 1 + 0.147 880 521 064 385 609 728;
  • 56) 0.147 880 521 064 385 609 728 × 2 = 0 + 0.295 761 042 128 771 219 456;
  • 57) 0.295 761 042 128 771 219 456 × 2 = 0 + 0.591 522 084 257 542 438 912;
  • 58) 0.591 522 084 257 542 438 912 × 2 = 1 + 0.183 044 168 515 084 877 824;
  • 59) 0.183 044 168 515 084 877 824 × 2 = 0 + 0.366 088 337 030 169 755 648;
  • 60) 0.366 088 337 030 169 755 648 × 2 = 0 + 0.732 176 674 060 339 511 296;
  • 61) 0.732 176 674 060 339 511 296 × 2 = 1 + 0.464 353 348 120 679 022 592;
  • 62) 0.464 353 348 120 679 022 592 × 2 = 0 + 0.928 706 696 241 358 045 184;
  • 63) 0.928 706 696 241 358 045 184 × 2 = 1 + 0.857 413 392 482 716 090 368;
  • 64) 0.857 413 392 482 716 090 368 × 2 = 1 + 0.714 826 784 965 432 180 736;
  • 65) 0.714 826 784 965 432 180 736 × 2 = 1 + 0.429 653 569 930 864 361 472;
  • 66) 0.429 653 569 930 864 361 472 × 2 = 0 + 0.859 307 139 861 728 722 944;
  • 67) 0.859 307 139 861 728 722 944 × 2 = 1 + 0.718 614 279 723 457 445 888;
  • 68) 0.718 614 279 723 457 445 888 × 2 = 1 + 0.437 228 559 446 914 891 776;
  • 69) 0.437 228 559 446 914 891 776 × 2 = 0 + 0.874 457 118 893 829 783 552;
  • 70) 0.874 457 118 893 829 783 552 × 2 = 1 + 0.748 914 237 787 659 567 104;
  • 71) 0.748 914 237 787 659 567 104 × 2 = 1 + 0.497 828 475 575 319 134 208;
  • 72) 0.497 828 475 575 319 134 208 × 2 = 0 + 0.995 656 951 150 638 268 416;
  • 73) 0.995 656 951 150 638 268 416 × 2 = 1 + 0.991 313 902 301 276 536 832;
  • 74) 0.991 313 902 301 276 536 832 × 2 = 1 + 0.982 627 804 602 553 073 664;
  • 75) 0.982 627 804 602 553 073 664 × 2 = 1 + 0.965 255 609 205 106 147 328;
  • 76) 0.965 255 609 205 106 147 328 × 2 = 1 + 0.930 511 218 410 212 294 656;
  • 77) 0.930 511 218 410 212 294 656 × 2 = 1 + 0.861 022 436 820 424 589 312;
  • 78) 0.861 022 436 820 424 589 312 × 2 = 1 + 0.722 044 873 640 849 178 624;
  • 79) 0.722 044 873 640 849 178 624 × 2 = 1 + 0.444 089 747 281 698 357 248;
  • 80) 0.444 089 747 281 698 357 248 × 2 = 0 + 0.888 179 494 563 396 714 496;
  • 81) 0.888 179 494 563 396 714 496 × 2 = 1 + 0.776 358 989 126 793 428 992;
  • 82) 0.776 358 989 126 793 428 992 × 2 = 1 + 0.552 717 978 253 586 857 984;
  • 83) 0.552 717 978 253 586 857 984 × 2 = 1 + 0.105 435 956 507 173 715 968;
  • 84) 0.105 435 956 507 173 715 968 × 2 = 0 + 0.210 871 913 014 347 431 936;
  • 85) 0.210 871 913 014 347 431 936 × 2 = 0 + 0.421 743 826 028 694 863 872;
  • 86) 0.421 743 826 028 694 863 872 × 2 = 0 + 0.843 487 652 057 389 727 744;
  • 87) 0.843 487 652 057 389 727 744 × 2 = 1 + 0.686 975 304 114 779 455 488;
  • 88) 0.686 975 304 114 779 455 488 × 2 = 1 + 0.373 950 608 229 558 910 976;
  • 89) 0.373 950 608 229 558 910 976 × 2 = 0 + 0.747 901 216 459 117 821 952;
  • 90) 0.747 901 216 459 117 821 952 × 2 = 1 + 0.495 802 432 918 235 643 904;
  • 91) 0.495 802 432 918 235 643 904 × 2 = 0 + 0.991 604 865 836 471 287 808;
  • 92) 0.991 604 865 836 471 287 808 × 2 = 1 + 0.983 209 731 672 942 575 616;
  • 93) 0.983 209 731 672 942 575 616 × 2 = 1 + 0.966 419 463 345 885 151 232;
  • 94) 0.966 419 463 345 885 151 232 × 2 = 1 + 0.932 838 926 691 770 302 464;
  • 95) 0.932 838 926 691 770 302 464 × 2 = 1 + 0.865 677 853 383 540 604 928;

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:


0.000 000 000 000 176 557 321(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1011 0110 1111 1110 1110 0011 0101 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 321(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1011 0110 1111 1110 1110 0011 0101 111(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:


0.000 000 000 000 176 557 321(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1011 0110 1111 1110 1110 0011 0101 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1011 0110 1111 1110 1110 0011 0101 111(2) × 20 =

1.1000 1101 1001 0010 0101 1101 1011 0111 1111 0111 0001 1010 1111(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 0101 1101 1011 0111 1111 0111 0001 1010 1111


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 0101 1101 1011 0111 1111 0111 0001 1010 1111 =

1000 1101 1001 0010 0101 1101 1011 0111 1111 0111 0001 1010 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 0100

Mantissa (52 bits) =
1000 1101 1001 0010 0101 1101 1011 0111 1111 0111 0001 1010 1111


Decimal number -0.000 000 000 000 176 557 321 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 1101 1011 0111 1111 0111 0001 1010 1111

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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