-0.000 000 000 000 176 557 477 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 477(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
-0.000 000 000 000 176 557 477(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:

|-0.000 000 000 000 176 557 477| = 0.000 000 000 000 176 557 477


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: 0.000 000 000 000 176 557 477.

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) 0.000 000 000 000 176 557 477 × 2 = 0 + 0.000 000 000 000 353 114 954;
  • 2) 0.000 000 000 000 353 114 954 × 2 = 0 + 0.000 000 000 000 706 229 908;
  • 3) 0.000 000 000 000 706 229 908 × 2 = 0 + 0.000 000 000 001 412 459 816;
  • 4) 0.000 000 000 001 412 459 816 × 2 = 0 + 0.000 000 000 002 824 919 632;
  • 5) 0.000 000 000 002 824 919 632 × 2 = 0 + 0.000 000 000 005 649 839 264;
  • 6) 0.000 000 000 005 649 839 264 × 2 = 0 + 0.000 000 000 011 299 678 528;
  • 7) 0.000 000 000 011 299 678 528 × 2 = 0 + 0.000 000 000 022 599 357 056;
  • 8) 0.000 000 000 022 599 357 056 × 2 = 0 + 0.000 000 000 045 198 714 112;
  • 9) 0.000 000 000 045 198 714 112 × 2 = 0 + 0.000 000 000 090 397 428 224;
  • 10) 0.000 000 000 090 397 428 224 × 2 = 0 + 0.000 000 000 180 794 856 448;
  • 11) 0.000 000 000 180 794 856 448 × 2 = 0 + 0.000 000 000 361 589 712 896;
  • 12) 0.000 000 000 361 589 712 896 × 2 = 0 + 0.000 000 000 723 179 425 792;
  • 13) 0.000 000 000 723 179 425 792 × 2 = 0 + 0.000 000 001 446 358 851 584;
  • 14) 0.000 000 001 446 358 851 584 × 2 = 0 + 0.000 000 002 892 717 703 168;
  • 15) 0.000 000 002 892 717 703 168 × 2 = 0 + 0.000 000 005 785 435 406 336;
  • 16) 0.000 000 005 785 435 406 336 × 2 = 0 + 0.000 000 011 570 870 812 672;
  • 17) 0.000 000 011 570 870 812 672 × 2 = 0 + 0.000 000 023 141 741 625 344;
  • 18) 0.000 000 023 141 741 625 344 × 2 = 0 + 0.000 000 046 283 483 250 688;
  • 19) 0.000 000 046 283 483 250 688 × 2 = 0 + 0.000 000 092 566 966 501 376;
  • 20) 0.000 000 092 566 966 501 376 × 2 = 0 + 0.000 000 185 133 933 002 752;
  • 21) 0.000 000 185 133 933 002 752 × 2 = 0 + 0.000 000 370 267 866 005 504;
  • 22) 0.000 000 370 267 866 005 504 × 2 = 0 + 0.000 000 740 535 732 011 008;
  • 23) 0.000 000 740 535 732 011 008 × 2 = 0 + 0.000 001 481 071 464 022 016;
  • 24) 0.000 001 481 071 464 022 016 × 2 = 0 + 0.000 002 962 142 928 044 032;
  • 25) 0.000 002 962 142 928 044 032 × 2 = 0 + 0.000 005 924 285 856 088 064;
  • 26) 0.000 005 924 285 856 088 064 × 2 = 0 + 0.000 011 848 571 712 176 128;
  • 27) 0.000 011 848 571 712 176 128 × 2 = 0 + 0.000 023 697 143 424 352 256;
  • 28) 0.000 023 697 143 424 352 256 × 2 = 0 + 0.000 047 394 286 848 704 512;
  • 29) 0.000 047 394 286 848 704 512 × 2 = 0 + 0.000 094 788 573 697 409 024;
  • 30) 0.000 094 788 573 697 409 024 × 2 = 0 + 0.000 189 577 147 394 818 048;
  • 31) 0.000 189 577 147 394 818 048 × 2 = 0 + 0.000 379 154 294 789 636 096;
  • 32) 0.000 379 154 294 789 636 096 × 2 = 0 + 0.000 758 308 589 579 272 192;
  • 33) 0.000 758 308 589 579 272 192 × 2 = 0 + 0.001 516 617 179 158 544 384;
  • 34) 0.001 516 617 179 158 544 384 × 2 = 0 + 0.003 033 234 358 317 088 768;
  • 35) 0.003 033 234 358 317 088 768 × 2 = 0 + 0.006 066 468 716 634 177 536;
  • 36) 0.006 066 468 716 634 177 536 × 2 = 0 + 0.012 132 937 433 268 355 072;
  • 37) 0.012 132 937 433 268 355 072 × 2 = 0 + 0.024 265 874 866 536 710 144;
  • 38) 0.024 265 874 866 536 710 144 × 2 = 0 + 0.048 531 749 733 073 420 288;
  • 39) 0.048 531 749 733 073 420 288 × 2 = 0 + 0.097 063 499 466 146 840 576;
  • 40) 0.097 063 499 466 146 840 576 × 2 = 0 + 0.194 126 998 932 293 681 152;
  • 41) 0.194 126 998 932 293 681 152 × 2 = 0 + 0.388 253 997 864 587 362 304;
  • 42) 0.388 253 997 864 587 362 304 × 2 = 0 + 0.776 507 995 729 174 724 608;
  • 43) 0.776 507 995 729 174 724 608 × 2 = 1 + 0.553 015 991 458 349 449 216;
  • 44) 0.553 015 991 458 349 449 216 × 2 = 1 + 0.106 031 982 916 698 898 432;
  • 45) 0.106 031 982 916 698 898 432 × 2 = 0 + 0.212 063 965 833 397 796 864;
  • 46) 0.212 063 965 833 397 796 864 × 2 = 0 + 0.424 127 931 666 795 593 728;
  • 47) 0.424 127 931 666 795 593 728 × 2 = 0 + 0.848 255 863 333 591 187 456;
  • 48) 0.848 255 863 333 591 187 456 × 2 = 1 + 0.696 511 726 667 182 374 912;
  • 49) 0.696 511 726 667 182 374 912 × 2 = 1 + 0.393 023 453 334 364 749 824;
  • 50) 0.393 023 453 334 364 749 824 × 2 = 0 + 0.786 046 906 668 729 499 648;
  • 51) 0.786 046 906 668 729 499 648 × 2 = 1 + 0.572 093 813 337 458 999 296;
  • 52) 0.572 093 813 337 458 999 296 × 2 = 1 + 0.144 187 626 674 917 998 592;
  • 53) 0.144 187 626 674 917 998 592 × 2 = 0 + 0.288 375 253 349 835 997 184;
  • 54) 0.288 375 253 349 835 997 184 × 2 = 0 + 0.576 750 506 699 671 994 368;
  • 55) 0.576 750 506 699 671 994 368 × 2 = 1 + 0.153 501 013 399 343 988 736;
  • 56) 0.153 501 013 399 343 988 736 × 2 = 0 + 0.307 002 026 798 687 977 472;
  • 57) 0.307 002 026 798 687 977 472 × 2 = 0 + 0.614 004 053 597 375 954 944;
  • 58) 0.614 004 053 597 375 954 944 × 2 = 1 + 0.228 008 107 194 751 909 888;
  • 59) 0.228 008 107 194 751 909 888 × 2 = 0 + 0.456 016 214 389 503 819 776;
  • 60) 0.456 016 214 389 503 819 776 × 2 = 0 + 0.912 032 428 779 007 639 552;
  • 61) 0.912 032 428 779 007 639 552 × 2 = 1 + 0.824 064 857 558 015 279 104;
  • 62) 0.824 064 857 558 015 279 104 × 2 = 1 + 0.648 129 715 116 030 558 208;
  • 63) 0.648 129 715 116 030 558 208 × 2 = 1 + 0.296 259 430 232 061 116 416;
  • 64) 0.296 259 430 232 061 116 416 × 2 = 0 + 0.592 518 860 464 122 232 832;
  • 65) 0.592 518 860 464 122 232 832 × 2 = 1 + 0.185 037 720 928 244 465 664;
  • 66) 0.185 037 720 928 244 465 664 × 2 = 0 + 0.370 075 441 856 488 931 328;
  • 67) 0.370 075 441 856 488 931 328 × 2 = 0 + 0.740 150 883 712 977 862 656;
  • 68) 0.740 150 883 712 977 862 656 × 2 = 1 + 0.480 301 767 425 955 725 312;
  • 69) 0.480 301 767 425 955 725 312 × 2 = 0 + 0.960 603 534 851 911 450 624;
  • 70) 0.960 603 534 851 911 450 624 × 2 = 1 + 0.921 207 069 703 822 901 248;
  • 71) 0.921 207 069 703 822 901 248 × 2 = 1 + 0.842 414 139 407 645 802 496;
  • 72) 0.842 414 139 407 645 802 496 × 2 = 1 + 0.684 828 278 815 291 604 992;
  • 73) 0.684 828 278 815 291 604 992 × 2 = 1 + 0.369 656 557 630 583 209 984;
  • 74) 0.369 656 557 630 583 209 984 × 2 = 0 + 0.739 313 115 261 166 419 968;
  • 75) 0.739 313 115 261 166 419 968 × 2 = 1 + 0.478 626 230 522 332 839 936;
  • 76) 0.478 626 230 522 332 839 936 × 2 = 0 + 0.957 252 461 044 665 679 872;
  • 77) 0.957 252 461 044 665 679 872 × 2 = 1 + 0.914 504 922 089 331 359 744;
  • 78) 0.914 504 922 089 331 359 744 × 2 = 1 + 0.829 009 844 178 662 719 488;
  • 79) 0.829 009 844 178 662 719 488 × 2 = 1 + 0.658 019 688 357 325 438 976;
  • 80) 0.658 019 688 357 325 438 976 × 2 = 1 + 0.316 039 376 714 650 877 952;
  • 81) 0.316 039 376 714 650 877 952 × 2 = 0 + 0.632 078 753 429 301 755 904;
  • 82) 0.632 078 753 429 301 755 904 × 2 = 1 + 0.264 157 506 858 603 511 808;
  • 83) 0.264 157 506 858 603 511 808 × 2 = 0 + 0.528 315 013 717 207 023 616;
  • 84) 0.528 315 013 717 207 023 616 × 2 = 1 + 0.056 630 027 434 414 047 232;
  • 85) 0.056 630 027 434 414 047 232 × 2 = 0 + 0.113 260 054 868 828 094 464;
  • 86) 0.113 260 054 868 828 094 464 × 2 = 0 + 0.226 520 109 737 656 188 928;
  • 87) 0.226 520 109 737 656 188 928 × 2 = 0 + 0.453 040 219 475 312 377 856;
  • 88) 0.453 040 219 475 312 377 856 × 2 = 0 + 0.906 080 438 950 624 755 712;
  • 89) 0.906 080 438 950 624 755 712 × 2 = 1 + 0.812 160 877 901 249 511 424;
  • 90) 0.812 160 877 901 249 511 424 × 2 = 1 + 0.624 321 755 802 499 022 848;
  • 91) 0.624 321 755 802 499 022 848 × 2 = 1 + 0.248 643 511 604 998 045 696;
  • 92) 0.248 643 511 604 998 045 696 × 2 = 0 + 0.497 287 023 209 996 091 392;
  • 93) 0.497 287 023 209 996 091 392 × 2 = 0 + 0.994 574 046 419 992 182 784;
  • 94) 0.994 574 046 419 992 182 784 × 2 = 1 + 0.989 148 092 839 984 365 568;
  • 95) 0.989 148 092 839 984 365 568 × 2 = 1 + 0.978 296 185 679 968 731 136;

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:


0.000 000 000 000 176 557 477(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1001 0111 1010 1111 0101 0000 1110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 477(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1001 0111 1010 1111 0101 0000 1110 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:


0.000 000 000 000 176 557 477(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1001 0111 1010 1111 0101 0000 1110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1001 0111 1010 1111 0101 0000 1110 011(2) × 20 =

1.1000 1101 1001 0010 0111 0100 1011 1101 0111 1010 1000 0111 0011(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 0100 1011 1101 0111 1010 1000 0111 0011


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 0100 1011 1101 0111 1010 1000 0111 0011 =

1000 1101 1001 0010 0111 0100 1011 1101 0111 1010 1000 0111 0011


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 0100 1011 1101 0111 1010 1000 0111 0011


Decimal number -0.000 000 000 000 176 557 477 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0111 0100 1011 1101 0111 1010 1000 0111 0011

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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