-0.000 000 000 000 176 557 483 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 483(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
-0.000 000 000 000 176 557 483(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:

|-0.000 000 000 000 176 557 483| = 0.000 000 000 000 176 557 483


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: 0.000 000 000 000 176 557 483.

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) 0.000 000 000 000 176 557 483 × 2 = 0 + 0.000 000 000 000 353 114 966;
  • 2) 0.000 000 000 000 353 114 966 × 2 = 0 + 0.000 000 000 000 706 229 932;
  • 3) 0.000 000 000 000 706 229 932 × 2 = 0 + 0.000 000 000 001 412 459 864;
  • 4) 0.000 000 000 001 412 459 864 × 2 = 0 + 0.000 000 000 002 824 919 728;
  • 5) 0.000 000 000 002 824 919 728 × 2 = 0 + 0.000 000 000 005 649 839 456;
  • 6) 0.000 000 000 005 649 839 456 × 2 = 0 + 0.000 000 000 011 299 678 912;
  • 7) 0.000 000 000 011 299 678 912 × 2 = 0 + 0.000 000 000 022 599 357 824;
  • 8) 0.000 000 000 022 599 357 824 × 2 = 0 + 0.000 000 000 045 198 715 648;
  • 9) 0.000 000 000 045 198 715 648 × 2 = 0 + 0.000 000 000 090 397 431 296;
  • 10) 0.000 000 000 090 397 431 296 × 2 = 0 + 0.000 000 000 180 794 862 592;
  • 11) 0.000 000 000 180 794 862 592 × 2 = 0 + 0.000 000 000 361 589 725 184;
  • 12) 0.000 000 000 361 589 725 184 × 2 = 0 + 0.000 000 000 723 179 450 368;
  • 13) 0.000 000 000 723 179 450 368 × 2 = 0 + 0.000 000 001 446 358 900 736;
  • 14) 0.000 000 001 446 358 900 736 × 2 = 0 + 0.000 000 002 892 717 801 472;
  • 15) 0.000 000 002 892 717 801 472 × 2 = 0 + 0.000 000 005 785 435 602 944;
  • 16) 0.000 000 005 785 435 602 944 × 2 = 0 + 0.000 000 011 570 871 205 888;
  • 17) 0.000 000 011 570 871 205 888 × 2 = 0 + 0.000 000 023 141 742 411 776;
  • 18) 0.000 000 023 141 742 411 776 × 2 = 0 + 0.000 000 046 283 484 823 552;
  • 19) 0.000 000 046 283 484 823 552 × 2 = 0 + 0.000 000 092 566 969 647 104;
  • 20) 0.000 000 092 566 969 647 104 × 2 = 0 + 0.000 000 185 133 939 294 208;
  • 21) 0.000 000 185 133 939 294 208 × 2 = 0 + 0.000 000 370 267 878 588 416;
  • 22) 0.000 000 370 267 878 588 416 × 2 = 0 + 0.000 000 740 535 757 176 832;
  • 23) 0.000 000 740 535 757 176 832 × 2 = 0 + 0.000 001 481 071 514 353 664;
  • 24) 0.000 001 481 071 514 353 664 × 2 = 0 + 0.000 002 962 143 028 707 328;
  • 25) 0.000 002 962 143 028 707 328 × 2 = 0 + 0.000 005 924 286 057 414 656;
  • 26) 0.000 005 924 286 057 414 656 × 2 = 0 + 0.000 011 848 572 114 829 312;
  • 27) 0.000 011 848 572 114 829 312 × 2 = 0 + 0.000 023 697 144 229 658 624;
  • 28) 0.000 023 697 144 229 658 624 × 2 = 0 + 0.000 047 394 288 459 317 248;
  • 29) 0.000 047 394 288 459 317 248 × 2 = 0 + 0.000 094 788 576 918 634 496;
  • 30) 0.000 094 788 576 918 634 496 × 2 = 0 + 0.000 189 577 153 837 268 992;
  • 31) 0.000 189 577 153 837 268 992 × 2 = 0 + 0.000 379 154 307 674 537 984;
  • 32) 0.000 379 154 307 674 537 984 × 2 = 0 + 0.000 758 308 615 349 075 968;
  • 33) 0.000 758 308 615 349 075 968 × 2 = 0 + 0.001 516 617 230 698 151 936;
  • 34) 0.001 516 617 230 698 151 936 × 2 = 0 + 0.003 033 234 461 396 303 872;
  • 35) 0.003 033 234 461 396 303 872 × 2 = 0 + 0.006 066 468 922 792 607 744;
  • 36) 0.006 066 468 922 792 607 744 × 2 = 0 + 0.012 132 937 845 585 215 488;
  • 37) 0.012 132 937 845 585 215 488 × 2 = 0 + 0.024 265 875 691 170 430 976;
  • 38) 0.024 265 875 691 170 430 976 × 2 = 0 + 0.048 531 751 382 340 861 952;
  • 39) 0.048 531 751 382 340 861 952 × 2 = 0 + 0.097 063 502 764 681 723 904;
  • 40) 0.097 063 502 764 681 723 904 × 2 = 0 + 0.194 127 005 529 363 447 808;
  • 41) 0.194 127 005 529 363 447 808 × 2 = 0 + 0.388 254 011 058 726 895 616;
  • 42) 0.388 254 011 058 726 895 616 × 2 = 0 + 0.776 508 022 117 453 791 232;
  • 43) 0.776 508 022 117 453 791 232 × 2 = 1 + 0.553 016 044 234 907 582 464;
  • 44) 0.553 016 044 234 907 582 464 × 2 = 1 + 0.106 032 088 469 815 164 928;
  • 45) 0.106 032 088 469 815 164 928 × 2 = 0 + 0.212 064 176 939 630 329 856;
  • 46) 0.212 064 176 939 630 329 856 × 2 = 0 + 0.424 128 353 879 260 659 712;
  • 47) 0.424 128 353 879 260 659 712 × 2 = 0 + 0.848 256 707 758 521 319 424;
  • 48) 0.848 256 707 758 521 319 424 × 2 = 1 + 0.696 513 415 517 042 638 848;
  • 49) 0.696 513 415 517 042 638 848 × 2 = 1 + 0.393 026 831 034 085 277 696;
  • 50) 0.393 026 831 034 085 277 696 × 2 = 0 + 0.786 053 662 068 170 555 392;
  • 51) 0.786 053 662 068 170 555 392 × 2 = 1 + 0.572 107 324 136 341 110 784;
  • 52) 0.572 107 324 136 341 110 784 × 2 = 1 + 0.144 214 648 272 682 221 568;
  • 53) 0.144 214 648 272 682 221 568 × 2 = 0 + 0.288 429 296 545 364 443 136;
  • 54) 0.288 429 296 545 364 443 136 × 2 = 0 + 0.576 858 593 090 728 886 272;
  • 55) 0.576 858 593 090 728 886 272 × 2 = 1 + 0.153 717 186 181 457 772 544;
  • 56) 0.153 717 186 181 457 772 544 × 2 = 0 + 0.307 434 372 362 915 545 088;
  • 57) 0.307 434 372 362 915 545 088 × 2 = 0 + 0.614 868 744 725 831 090 176;
  • 58) 0.614 868 744 725 831 090 176 × 2 = 1 + 0.229 737 489 451 662 180 352;
  • 59) 0.229 737 489 451 662 180 352 × 2 = 0 + 0.459 474 978 903 324 360 704;
  • 60) 0.459 474 978 903 324 360 704 × 2 = 0 + 0.918 949 957 806 648 721 408;
  • 61) 0.918 949 957 806 648 721 408 × 2 = 1 + 0.837 899 915 613 297 442 816;
  • 62) 0.837 899 915 613 297 442 816 × 2 = 1 + 0.675 799 831 226 594 885 632;
  • 63) 0.675 799 831 226 594 885 632 × 2 = 1 + 0.351 599 662 453 189 771 264;
  • 64) 0.351 599 662 453 189 771 264 × 2 = 0 + 0.703 199 324 906 379 542 528;
  • 65) 0.703 199 324 906 379 542 528 × 2 = 1 + 0.406 398 649 812 759 085 056;
  • 66) 0.406 398 649 812 759 085 056 × 2 = 0 + 0.812 797 299 625 518 170 112;
  • 67) 0.812 797 299 625 518 170 112 × 2 = 1 + 0.625 594 599 251 036 340 224;
  • 68) 0.625 594 599 251 036 340 224 × 2 = 1 + 0.251 189 198 502 072 680 448;
  • 69) 0.251 189 198 502 072 680 448 × 2 = 0 + 0.502 378 397 004 145 360 896;
  • 70) 0.502 378 397 004 145 360 896 × 2 = 1 + 0.004 756 794 008 290 721 792;
  • 71) 0.004 756 794 008 290 721 792 × 2 = 0 + 0.009 513 588 016 581 443 584;
  • 72) 0.009 513 588 016 581 443 584 × 2 = 0 + 0.019 027 176 033 162 887 168;
  • 73) 0.019 027 176 033 162 887 168 × 2 = 0 + 0.038 054 352 066 325 774 336;
  • 74) 0.038 054 352 066 325 774 336 × 2 = 0 + 0.076 108 704 132 651 548 672;
  • 75) 0.076 108 704 132 651 548 672 × 2 = 0 + 0.152 217 408 265 303 097 344;
  • 76) 0.152 217 408 265 303 097 344 × 2 = 0 + 0.304 434 816 530 606 194 688;
  • 77) 0.304 434 816 530 606 194 688 × 2 = 0 + 0.608 869 633 061 212 389 376;
  • 78) 0.608 869 633 061 212 389 376 × 2 = 1 + 0.217 739 266 122 424 778 752;
  • 79) 0.217 739 266 122 424 778 752 × 2 = 0 + 0.435 478 532 244 849 557 504;
  • 80) 0.435 478 532 244 849 557 504 × 2 = 0 + 0.870 957 064 489 699 115 008;
  • 81) 0.870 957 064 489 699 115 008 × 2 = 1 + 0.741 914 128 979 398 230 016;
  • 82) 0.741 914 128 979 398 230 016 × 2 = 1 + 0.483 828 257 958 796 460 032;
  • 83) 0.483 828 257 958 796 460 032 × 2 = 0 + 0.967 656 515 917 592 920 064;
  • 84) 0.967 656 515 917 592 920 064 × 2 = 1 + 0.935 313 031 835 185 840 128;
  • 85) 0.935 313 031 835 185 840 128 × 2 = 1 + 0.870 626 063 670 371 680 256;
  • 86) 0.870 626 063 670 371 680 256 × 2 = 1 + 0.741 252 127 340 743 360 512;
  • 87) 0.741 252 127 340 743 360 512 × 2 = 1 + 0.482 504 254 681 486 721 024;
  • 88) 0.482 504 254 681 486 721 024 × 2 = 0 + 0.965 008 509 362 973 442 048;
  • 89) 0.965 008 509 362 973 442 048 × 2 = 1 + 0.930 017 018 725 946 884 096;
  • 90) 0.930 017 018 725 946 884 096 × 2 = 1 + 0.860 034 037 451 893 768 192;
  • 91) 0.860 034 037 451 893 768 192 × 2 = 1 + 0.720 068 074 903 787 536 384;
  • 92) 0.720 068 074 903 787 536 384 × 2 = 1 + 0.440 136 149 807 575 072 768;
  • 93) 0.440 136 149 807 575 072 768 × 2 = 0 + 0.880 272 299 615 150 145 536;
  • 94) 0.880 272 299 615 150 145 536 × 2 = 1 + 0.760 544 599 230 300 291 072;
  • 95) 0.760 544 599 230 300 291 072 × 2 = 1 + 0.521 089 198 460 600 582 144;

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:


0.000 000 000 000 176 557 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1011 0100 0000 0100 1101 1110 1111 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1011 0100 0000 0100 1101 1110 1111 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:


0.000 000 000 000 176 557 483(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1011 0100 0000 0100 1101 1110 1111 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1110 1011 0100 0000 0100 1101 1110 1111 011(2) × 20 =

1.1000 1101 1001 0010 0111 0101 1010 0000 0010 0110 1111 0111 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 0101 1010 0000 0010 0110 1111 0111 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 0101 1010 0000 0010 0110 1111 0111 1011 =

1000 1101 1001 0010 0111 0101 1010 0000 0010 0110 1111 0111 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 0101 1010 0000 0010 0110 1111 0111 1011


Decimal number -0.000 000 000 000 176 557 483 converted to 64 bit double precision IEEE 754 binary floating point representation:

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