-0.000 000 000 000 176 557 334 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 334(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
-0.000 000 000 000 176 557 334(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:

|-0.000 000 000 000 176 557 334| = 0.000 000 000 000 176 557 334


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: 0.000 000 000 000 176 557 334.

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) 0.000 000 000 000 176 557 334 × 2 = 0 + 0.000 000 000 000 353 114 668;
  • 2) 0.000 000 000 000 353 114 668 × 2 = 0 + 0.000 000 000 000 706 229 336;
  • 3) 0.000 000 000 000 706 229 336 × 2 = 0 + 0.000 000 000 001 412 458 672;
  • 4) 0.000 000 000 001 412 458 672 × 2 = 0 + 0.000 000 000 002 824 917 344;
  • 5) 0.000 000 000 002 824 917 344 × 2 = 0 + 0.000 000 000 005 649 834 688;
  • 6) 0.000 000 000 005 649 834 688 × 2 = 0 + 0.000 000 000 011 299 669 376;
  • 7) 0.000 000 000 011 299 669 376 × 2 = 0 + 0.000 000 000 022 599 338 752;
  • 8) 0.000 000 000 022 599 338 752 × 2 = 0 + 0.000 000 000 045 198 677 504;
  • 9) 0.000 000 000 045 198 677 504 × 2 = 0 + 0.000 000 000 090 397 355 008;
  • 10) 0.000 000 000 090 397 355 008 × 2 = 0 + 0.000 000 000 180 794 710 016;
  • 11) 0.000 000 000 180 794 710 016 × 2 = 0 + 0.000 000 000 361 589 420 032;
  • 12) 0.000 000 000 361 589 420 032 × 2 = 0 + 0.000 000 000 723 178 840 064;
  • 13) 0.000 000 000 723 178 840 064 × 2 = 0 + 0.000 000 001 446 357 680 128;
  • 14) 0.000 000 001 446 357 680 128 × 2 = 0 + 0.000 000 002 892 715 360 256;
  • 15) 0.000 000 002 892 715 360 256 × 2 = 0 + 0.000 000 005 785 430 720 512;
  • 16) 0.000 000 005 785 430 720 512 × 2 = 0 + 0.000 000 011 570 861 441 024;
  • 17) 0.000 000 011 570 861 441 024 × 2 = 0 + 0.000 000 023 141 722 882 048;
  • 18) 0.000 000 023 141 722 882 048 × 2 = 0 + 0.000 000 046 283 445 764 096;
  • 19) 0.000 000 046 283 445 764 096 × 2 = 0 + 0.000 000 092 566 891 528 192;
  • 20) 0.000 000 092 566 891 528 192 × 2 = 0 + 0.000 000 185 133 783 056 384;
  • 21) 0.000 000 185 133 783 056 384 × 2 = 0 + 0.000 000 370 267 566 112 768;
  • 22) 0.000 000 370 267 566 112 768 × 2 = 0 + 0.000 000 740 535 132 225 536;
  • 23) 0.000 000 740 535 132 225 536 × 2 = 0 + 0.000 001 481 070 264 451 072;
  • 24) 0.000 001 481 070 264 451 072 × 2 = 0 + 0.000 002 962 140 528 902 144;
  • 25) 0.000 002 962 140 528 902 144 × 2 = 0 + 0.000 005 924 281 057 804 288;
  • 26) 0.000 005 924 281 057 804 288 × 2 = 0 + 0.000 011 848 562 115 608 576;
  • 27) 0.000 011 848 562 115 608 576 × 2 = 0 + 0.000 023 697 124 231 217 152;
  • 28) 0.000 023 697 124 231 217 152 × 2 = 0 + 0.000 047 394 248 462 434 304;
  • 29) 0.000 047 394 248 462 434 304 × 2 = 0 + 0.000 094 788 496 924 868 608;
  • 30) 0.000 094 788 496 924 868 608 × 2 = 0 + 0.000 189 576 993 849 737 216;
  • 31) 0.000 189 576 993 849 737 216 × 2 = 0 + 0.000 379 153 987 699 474 432;
  • 32) 0.000 379 153 987 699 474 432 × 2 = 0 + 0.000 758 307 975 398 948 864;
  • 33) 0.000 758 307 975 398 948 864 × 2 = 0 + 0.001 516 615 950 797 897 728;
  • 34) 0.001 516 615 950 797 897 728 × 2 = 0 + 0.003 033 231 901 595 795 456;
  • 35) 0.003 033 231 901 595 795 456 × 2 = 0 + 0.006 066 463 803 191 590 912;
  • 36) 0.006 066 463 803 191 590 912 × 2 = 0 + 0.012 132 927 606 383 181 824;
  • 37) 0.012 132 927 606 383 181 824 × 2 = 0 + 0.024 265 855 212 766 363 648;
  • 38) 0.024 265 855 212 766 363 648 × 2 = 0 + 0.048 531 710 425 532 727 296;
  • 39) 0.048 531 710 425 532 727 296 × 2 = 0 + 0.097 063 420 851 065 454 592;
  • 40) 0.097 063 420 851 065 454 592 × 2 = 0 + 0.194 126 841 702 130 909 184;
  • 41) 0.194 126 841 702 130 909 184 × 2 = 0 + 0.388 253 683 404 261 818 368;
  • 42) 0.388 253 683 404 261 818 368 × 2 = 0 + 0.776 507 366 808 523 636 736;
  • 43) 0.776 507 366 808 523 636 736 × 2 = 1 + 0.553 014 733 617 047 273 472;
  • 44) 0.553 014 733 617 047 273 472 × 2 = 1 + 0.106 029 467 234 094 546 944;
  • 45) 0.106 029 467 234 094 546 944 × 2 = 0 + 0.212 058 934 468 189 093 888;
  • 46) 0.212 058 934 468 189 093 888 × 2 = 0 + 0.424 117 868 936 378 187 776;
  • 47) 0.424 117 868 936 378 187 776 × 2 = 0 + 0.848 235 737 872 756 375 552;
  • 48) 0.848 235 737 872 756 375 552 × 2 = 1 + 0.696 471 475 745 512 751 104;
  • 49) 0.696 471 475 745 512 751 104 × 2 = 1 + 0.392 942 951 491 025 502 208;
  • 50) 0.392 942 951 491 025 502 208 × 2 = 0 + 0.785 885 902 982 051 004 416;
  • 51) 0.785 885 902 982 051 004 416 × 2 = 1 + 0.571 771 805 964 102 008 832;
  • 52) 0.571 771 805 964 102 008 832 × 2 = 1 + 0.143 543 611 928 204 017 664;
  • 53) 0.143 543 611 928 204 017 664 × 2 = 0 + 0.287 087 223 856 408 035 328;
  • 54) 0.287 087 223 856 408 035 328 × 2 = 0 + 0.574 174 447 712 816 070 656;
  • 55) 0.574 174 447 712 816 070 656 × 2 = 1 + 0.148 348 895 425 632 141 312;
  • 56) 0.148 348 895 425 632 141 312 × 2 = 0 + 0.296 697 790 851 264 282 624;
  • 57) 0.296 697 790 851 264 282 624 × 2 = 0 + 0.593 395 581 702 528 565 248;
  • 58) 0.593 395 581 702 528 565 248 × 2 = 1 + 0.186 791 163 405 057 130 496;
  • 59) 0.186 791 163 405 057 130 496 × 2 = 0 + 0.373 582 326 810 114 260 992;
  • 60) 0.373 582 326 810 114 260 992 × 2 = 0 + 0.747 164 653 620 228 521 984;
  • 61) 0.747 164 653 620 228 521 984 × 2 = 1 + 0.494 329 307 240 457 043 968;
  • 62) 0.494 329 307 240 457 043 968 × 2 = 0 + 0.988 658 614 480 914 087 936;
  • 63) 0.988 658 614 480 914 087 936 × 2 = 1 + 0.977 317 228 961 828 175 872;
  • 64) 0.977 317 228 961 828 175 872 × 2 = 1 + 0.954 634 457 923 656 351 744;
  • 65) 0.954 634 457 923 656 351 744 × 2 = 1 + 0.909 268 915 847 312 703 488;
  • 66) 0.909 268 915 847 312 703 488 × 2 = 1 + 0.818 537 831 694 625 406 976;
  • 67) 0.818 537 831 694 625 406 976 × 2 = 1 + 0.637 075 663 389 250 813 952;
  • 68) 0.637 075 663 389 250 813 952 × 2 = 1 + 0.274 151 326 778 501 627 904;
  • 69) 0.274 151 326 778 501 627 904 × 2 = 0 + 0.548 302 653 557 003 255 808;
  • 70) 0.548 302 653 557 003 255 808 × 2 = 1 + 0.096 605 307 114 006 511 616;
  • 71) 0.096 605 307 114 006 511 616 × 2 = 0 + 0.193 210 614 228 013 023 232;
  • 72) 0.193 210 614 228 013 023 232 × 2 = 0 + 0.386 421 228 456 026 046 464;
  • 73) 0.386 421 228 456 026 046 464 × 2 = 0 + 0.772 842 456 912 052 092 928;
  • 74) 0.772 842 456 912 052 092 928 × 2 = 1 + 0.545 684 913 824 104 185 856;
  • 75) 0.545 684 913 824 104 185 856 × 2 = 1 + 0.091 369 827 648 208 371 712;
  • 76) 0.091 369 827 648 208 371 712 × 2 = 0 + 0.182 739 655 296 416 743 424;
  • 77) 0.182 739 655 296 416 743 424 × 2 = 0 + 0.365 479 310 592 833 486 848;
  • 78) 0.365 479 310 592 833 486 848 × 2 = 0 + 0.730 958 621 185 666 973 696;
  • 79) 0.730 958 621 185 666 973 696 × 2 = 1 + 0.461 917 242 371 333 947 392;
  • 80) 0.461 917 242 371 333 947 392 × 2 = 0 + 0.923 834 484 742 667 894 784;
  • 81) 0.923 834 484 742 667 894 784 × 2 = 1 + 0.847 668 969 485 335 789 568;
  • 82) 0.847 668 969 485 335 789 568 × 2 = 1 + 0.695 337 938 970 671 579 136;
  • 83) 0.695 337 938 970 671 579 136 × 2 = 1 + 0.390 675 877 941 343 158 272;
  • 84) 0.390 675 877 941 343 158 272 × 2 = 0 + 0.781 351 755 882 686 316 544;
  • 85) 0.781 351 755 882 686 316 544 × 2 = 1 + 0.562 703 511 765 372 633 088;
  • 86) 0.562 703 511 765 372 633 088 × 2 = 1 + 0.125 407 023 530 745 266 176;
  • 87) 0.125 407 023 530 745 266 176 × 2 = 0 + 0.250 814 047 061 490 532 352;
  • 88) 0.250 814 047 061 490 532 352 × 2 = 0 + 0.501 628 094 122 981 064 704;
  • 89) 0.501 628 094 122 981 064 704 × 2 = 1 + 0.003 256 188 245 962 129 408;
  • 90) 0.003 256 188 245 962 129 408 × 2 = 0 + 0.006 512 376 491 924 258 816;
  • 91) 0.006 512 376 491 924 258 816 × 2 = 0 + 0.013 024 752 983 848 517 632;
  • 92) 0.013 024 752 983 848 517 632 × 2 = 0 + 0.026 049 505 967 697 035 264;
  • 93) 0.026 049 505 967 697 035 264 × 2 = 0 + 0.052 099 011 935 394 070 528;
  • 94) 0.052 099 011 935 394 070 528 × 2 = 0 + 0.104 198 023 870 788 141 056;
  • 95) 0.104 198 023 870 788 141 056 × 2 = 0 + 0.208 396 047 741 576 282 112;

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:


0.000 000 000 000 176 557 334(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1111 0100 0110 0010 1110 1100 1000 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 334(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1111 0100 0110 0010 1110 1100 1000 000(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:


0.000 000 000 000 176 557 334(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1111 0100 0110 0010 1110 1100 1000 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1111 0100 0110 0010 1110 1100 1000 000(2) × 20 =

1.1000 1101 1001 0010 0101 1111 1010 0011 0001 0111 0110 0100 0000(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 1111 1010 0011 0001 0111 0110 0100 0000


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 1111 1010 0011 0001 0111 0110 0100 0000 =

1000 1101 1001 0010 0101 1111 1010 0011 0001 0111 0110 0100 0000


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 1111 1010 0011 0001 0111 0110 0100 0000


Decimal number -0.000 000 000 000 176 557 334 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 1111 1010 0011 0001 0111 0110 0100 0000

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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