-0.000 000 000 000 176 557 388 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 388(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
-0.000 000 000 000 176 557 388(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:

|-0.000 000 000 000 176 557 388| = 0.000 000 000 000 176 557 388


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: 0.000 000 000 000 176 557 388.

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) 0.000 000 000 000 176 557 388 × 2 = 0 + 0.000 000 000 000 353 114 776;
  • 2) 0.000 000 000 000 353 114 776 × 2 = 0 + 0.000 000 000 000 706 229 552;
  • 3) 0.000 000 000 000 706 229 552 × 2 = 0 + 0.000 000 000 001 412 459 104;
  • 4) 0.000 000 000 001 412 459 104 × 2 = 0 + 0.000 000 000 002 824 918 208;
  • 5) 0.000 000 000 002 824 918 208 × 2 = 0 + 0.000 000 000 005 649 836 416;
  • 6) 0.000 000 000 005 649 836 416 × 2 = 0 + 0.000 000 000 011 299 672 832;
  • 7) 0.000 000 000 011 299 672 832 × 2 = 0 + 0.000 000 000 022 599 345 664;
  • 8) 0.000 000 000 022 599 345 664 × 2 = 0 + 0.000 000 000 045 198 691 328;
  • 9) 0.000 000 000 045 198 691 328 × 2 = 0 + 0.000 000 000 090 397 382 656;
  • 10) 0.000 000 000 090 397 382 656 × 2 = 0 + 0.000 000 000 180 794 765 312;
  • 11) 0.000 000 000 180 794 765 312 × 2 = 0 + 0.000 000 000 361 589 530 624;
  • 12) 0.000 000 000 361 589 530 624 × 2 = 0 + 0.000 000 000 723 179 061 248;
  • 13) 0.000 000 000 723 179 061 248 × 2 = 0 + 0.000 000 001 446 358 122 496;
  • 14) 0.000 000 001 446 358 122 496 × 2 = 0 + 0.000 000 002 892 716 244 992;
  • 15) 0.000 000 002 892 716 244 992 × 2 = 0 + 0.000 000 005 785 432 489 984;
  • 16) 0.000 000 005 785 432 489 984 × 2 = 0 + 0.000 000 011 570 864 979 968;
  • 17) 0.000 000 011 570 864 979 968 × 2 = 0 + 0.000 000 023 141 729 959 936;
  • 18) 0.000 000 023 141 729 959 936 × 2 = 0 + 0.000 000 046 283 459 919 872;
  • 19) 0.000 000 046 283 459 919 872 × 2 = 0 + 0.000 000 092 566 919 839 744;
  • 20) 0.000 000 092 566 919 839 744 × 2 = 0 + 0.000 000 185 133 839 679 488;
  • 21) 0.000 000 185 133 839 679 488 × 2 = 0 + 0.000 000 370 267 679 358 976;
  • 22) 0.000 000 370 267 679 358 976 × 2 = 0 + 0.000 000 740 535 358 717 952;
  • 23) 0.000 000 740 535 358 717 952 × 2 = 0 + 0.000 001 481 070 717 435 904;
  • 24) 0.000 001 481 070 717 435 904 × 2 = 0 + 0.000 002 962 141 434 871 808;
  • 25) 0.000 002 962 141 434 871 808 × 2 = 0 + 0.000 005 924 282 869 743 616;
  • 26) 0.000 005 924 282 869 743 616 × 2 = 0 + 0.000 011 848 565 739 487 232;
  • 27) 0.000 011 848 565 739 487 232 × 2 = 0 + 0.000 023 697 131 478 974 464;
  • 28) 0.000 023 697 131 478 974 464 × 2 = 0 + 0.000 047 394 262 957 948 928;
  • 29) 0.000 047 394 262 957 948 928 × 2 = 0 + 0.000 094 788 525 915 897 856;
  • 30) 0.000 094 788 525 915 897 856 × 2 = 0 + 0.000 189 577 051 831 795 712;
  • 31) 0.000 189 577 051 831 795 712 × 2 = 0 + 0.000 379 154 103 663 591 424;
  • 32) 0.000 379 154 103 663 591 424 × 2 = 0 + 0.000 758 308 207 327 182 848;
  • 33) 0.000 758 308 207 327 182 848 × 2 = 0 + 0.001 516 616 414 654 365 696;
  • 34) 0.001 516 616 414 654 365 696 × 2 = 0 + 0.003 033 232 829 308 731 392;
  • 35) 0.003 033 232 829 308 731 392 × 2 = 0 + 0.006 066 465 658 617 462 784;
  • 36) 0.006 066 465 658 617 462 784 × 2 = 0 + 0.012 132 931 317 234 925 568;
  • 37) 0.012 132 931 317 234 925 568 × 2 = 0 + 0.024 265 862 634 469 851 136;
  • 38) 0.024 265 862 634 469 851 136 × 2 = 0 + 0.048 531 725 268 939 702 272;
  • 39) 0.048 531 725 268 939 702 272 × 2 = 0 + 0.097 063 450 537 879 404 544;
  • 40) 0.097 063 450 537 879 404 544 × 2 = 0 + 0.194 126 901 075 758 809 088;
  • 41) 0.194 126 901 075 758 809 088 × 2 = 0 + 0.388 253 802 151 517 618 176;
  • 42) 0.388 253 802 151 517 618 176 × 2 = 0 + 0.776 507 604 303 035 236 352;
  • 43) 0.776 507 604 303 035 236 352 × 2 = 1 + 0.553 015 208 606 070 472 704;
  • 44) 0.553 015 208 606 070 472 704 × 2 = 1 + 0.106 030 417 212 140 945 408;
  • 45) 0.106 030 417 212 140 945 408 × 2 = 0 + 0.212 060 834 424 281 890 816;
  • 46) 0.212 060 834 424 281 890 816 × 2 = 0 + 0.424 121 668 848 563 781 632;
  • 47) 0.424 121 668 848 563 781 632 × 2 = 0 + 0.848 243 337 697 127 563 264;
  • 48) 0.848 243 337 697 127 563 264 × 2 = 1 + 0.696 486 675 394 255 126 528;
  • 49) 0.696 486 675 394 255 126 528 × 2 = 1 + 0.392 973 350 788 510 253 056;
  • 50) 0.392 973 350 788 510 253 056 × 2 = 0 + 0.785 946 701 577 020 506 112;
  • 51) 0.785 946 701 577 020 506 112 × 2 = 1 + 0.571 893 403 154 041 012 224;
  • 52) 0.571 893 403 154 041 012 224 × 2 = 1 + 0.143 786 806 308 082 024 448;
  • 53) 0.143 786 806 308 082 024 448 × 2 = 0 + 0.287 573 612 616 164 048 896;
  • 54) 0.287 573 612 616 164 048 896 × 2 = 0 + 0.575 147 225 232 328 097 792;
  • 55) 0.575 147 225 232 328 097 792 × 2 = 1 + 0.150 294 450 464 656 195 584;
  • 56) 0.150 294 450 464 656 195 584 × 2 = 0 + 0.300 588 900 929 312 391 168;
  • 57) 0.300 588 900 929 312 391 168 × 2 = 0 + 0.601 177 801 858 624 782 336;
  • 58) 0.601 177 801 858 624 782 336 × 2 = 1 + 0.202 355 603 717 249 564 672;
  • 59) 0.202 355 603 717 249 564 672 × 2 = 0 + 0.404 711 207 434 499 129 344;
  • 60) 0.404 711 207 434 499 129 344 × 2 = 0 + 0.809 422 414 868 998 258 688;
  • 61) 0.809 422 414 868 998 258 688 × 2 = 1 + 0.618 844 829 737 996 517 376;
  • 62) 0.618 844 829 737 996 517 376 × 2 = 1 + 0.237 689 659 475 993 034 752;
  • 63) 0.237 689 659 475 993 034 752 × 2 = 0 + 0.475 379 318 951 986 069 504;
  • 64) 0.475 379 318 951 986 069 504 × 2 = 0 + 0.950 758 637 903 972 139 008;
  • 65) 0.950 758 637 903 972 139 008 × 2 = 1 + 0.901 517 275 807 944 278 016;
  • 66) 0.901 517 275 807 944 278 016 × 2 = 1 + 0.803 034 551 615 888 556 032;
  • 67) 0.803 034 551 615 888 556 032 × 2 = 1 + 0.606 069 103 231 777 112 064;
  • 68) 0.606 069 103 231 777 112 064 × 2 = 1 + 0.212 138 206 463 554 224 128;
  • 69) 0.212 138 206 463 554 224 128 × 2 = 0 + 0.424 276 412 927 108 448 256;
  • 70) 0.424 276 412 927 108 448 256 × 2 = 0 + 0.848 552 825 854 216 896 512;
  • 71) 0.848 552 825 854 216 896 512 × 2 = 1 + 0.697 105 651 708 433 793 024;
  • 72) 0.697 105 651 708 433 793 024 × 2 = 1 + 0.394 211 303 416 867 586 048;
  • 73) 0.394 211 303 416 867 586 048 × 2 = 0 + 0.788 422 606 833 735 172 096;
  • 74) 0.788 422 606 833 735 172 096 × 2 = 1 + 0.576 845 213 667 470 344 192;
  • 75) 0.576 845 213 667 470 344 192 × 2 = 1 + 0.153 690 427 334 940 688 384;
  • 76) 0.153 690 427 334 940 688 384 × 2 = 0 + 0.307 380 854 669 881 376 768;
  • 77) 0.307 380 854 669 881 376 768 × 2 = 0 + 0.614 761 709 339 762 753 536;
  • 78) 0.614 761 709 339 762 753 536 × 2 = 1 + 0.229 523 418 679 525 507 072;
  • 79) 0.229 523 418 679 525 507 072 × 2 = 0 + 0.459 046 837 359 051 014 144;
  • 80) 0.459 046 837 359 051 014 144 × 2 = 0 + 0.918 093 674 718 102 028 288;
  • 81) 0.918 093 674 718 102 028 288 × 2 = 1 + 0.836 187 349 436 204 056 576;
  • 82) 0.836 187 349 436 204 056 576 × 2 = 1 + 0.672 374 698 872 408 113 152;
  • 83) 0.672 374 698 872 408 113 152 × 2 = 1 + 0.344 749 397 744 816 226 304;
  • 84) 0.344 749 397 744 816 226 304 × 2 = 0 + 0.689 498 795 489 632 452 608;
  • 85) 0.689 498 795 489 632 452 608 × 2 = 1 + 0.378 997 590 979 264 905 216;
  • 86) 0.378 997 590 979 264 905 216 × 2 = 0 + 0.757 995 181 958 529 810 432;
  • 87) 0.757 995 181 958 529 810 432 × 2 = 1 + 0.515 990 363 917 059 620 864;
  • 88) 0.515 990 363 917 059 620 864 × 2 = 1 + 0.031 980 727 834 119 241 728;
  • 89) 0.031 980 727 834 119 241 728 × 2 = 0 + 0.063 961 455 668 238 483 456;
  • 90) 0.063 961 455 668 238 483 456 × 2 = 0 + 0.127 922 911 336 476 966 912;
  • 91) 0.127 922 911 336 476 966 912 × 2 = 0 + 0.255 845 822 672 953 933 824;
  • 92) 0.255 845 822 672 953 933 824 × 2 = 0 + 0.511 691 645 345 907 867 648;
  • 93) 0.511 691 645 345 907 867 648 × 2 = 1 + 0.023 383 290 691 815 735 296;
  • 94) 0.023 383 290 691 815 735 296 × 2 = 0 + 0.046 766 581 383 631 470 592;
  • 95) 0.046 766 581 383 631 470 592 × 2 = 0 + 0.093 533 162 767 262 941 184;

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:


0.000 000 000 000 176 557 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1111 0011 0110 0100 1110 1011 0000 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1111 0011 0110 0100 1110 1011 0000 100(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:


0.000 000 000 000 176 557 388(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1111 0011 0110 0100 1110 1011 0000 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1100 1111 0011 0110 0100 1110 1011 0000 100(2) × 20 =

1.1000 1101 1001 0010 0110 0111 1001 1011 0010 0111 0101 1000 0100(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 0110 0111 1001 1011 0010 0111 0101 1000 0100


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 0110 0111 1001 1011 0010 0111 0101 1000 0100 =

1000 1101 1001 0010 0110 0111 1001 1011 0010 0111 0101 1000 0100


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 0110 0111 1001 1011 0010 0111 0101 1000 0100


Decimal number -0.000 000 000 000 176 557 388 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0110 0111 1001 1011 0010 0111 0101 1000 0100

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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