-0.000 000 000 000 176 557 254 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 254(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
-0.000 000 000 000 176 557 254(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:

|-0.000 000 000 000 176 557 254| = 0.000 000 000 000 176 557 254


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: 0.000 000 000 000 176 557 254.

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) 0.000 000 000 000 176 557 254 × 2 = 0 + 0.000 000 000 000 353 114 508;
  • 2) 0.000 000 000 000 353 114 508 × 2 = 0 + 0.000 000 000 000 706 229 016;
  • 3) 0.000 000 000 000 706 229 016 × 2 = 0 + 0.000 000 000 001 412 458 032;
  • 4) 0.000 000 000 001 412 458 032 × 2 = 0 + 0.000 000 000 002 824 916 064;
  • 5) 0.000 000 000 002 824 916 064 × 2 = 0 + 0.000 000 000 005 649 832 128;
  • 6) 0.000 000 000 005 649 832 128 × 2 = 0 + 0.000 000 000 011 299 664 256;
  • 7) 0.000 000 000 011 299 664 256 × 2 = 0 + 0.000 000 000 022 599 328 512;
  • 8) 0.000 000 000 022 599 328 512 × 2 = 0 + 0.000 000 000 045 198 657 024;
  • 9) 0.000 000 000 045 198 657 024 × 2 = 0 + 0.000 000 000 090 397 314 048;
  • 10) 0.000 000 000 090 397 314 048 × 2 = 0 + 0.000 000 000 180 794 628 096;
  • 11) 0.000 000 000 180 794 628 096 × 2 = 0 + 0.000 000 000 361 589 256 192;
  • 12) 0.000 000 000 361 589 256 192 × 2 = 0 + 0.000 000 000 723 178 512 384;
  • 13) 0.000 000 000 723 178 512 384 × 2 = 0 + 0.000 000 001 446 357 024 768;
  • 14) 0.000 000 001 446 357 024 768 × 2 = 0 + 0.000 000 002 892 714 049 536;
  • 15) 0.000 000 002 892 714 049 536 × 2 = 0 + 0.000 000 005 785 428 099 072;
  • 16) 0.000 000 005 785 428 099 072 × 2 = 0 + 0.000 000 011 570 856 198 144;
  • 17) 0.000 000 011 570 856 198 144 × 2 = 0 + 0.000 000 023 141 712 396 288;
  • 18) 0.000 000 023 141 712 396 288 × 2 = 0 + 0.000 000 046 283 424 792 576;
  • 19) 0.000 000 046 283 424 792 576 × 2 = 0 + 0.000 000 092 566 849 585 152;
  • 20) 0.000 000 092 566 849 585 152 × 2 = 0 + 0.000 000 185 133 699 170 304;
  • 21) 0.000 000 185 133 699 170 304 × 2 = 0 + 0.000 000 370 267 398 340 608;
  • 22) 0.000 000 370 267 398 340 608 × 2 = 0 + 0.000 000 740 534 796 681 216;
  • 23) 0.000 000 740 534 796 681 216 × 2 = 0 + 0.000 001 481 069 593 362 432;
  • 24) 0.000 001 481 069 593 362 432 × 2 = 0 + 0.000 002 962 139 186 724 864;
  • 25) 0.000 002 962 139 186 724 864 × 2 = 0 + 0.000 005 924 278 373 449 728;
  • 26) 0.000 005 924 278 373 449 728 × 2 = 0 + 0.000 011 848 556 746 899 456;
  • 27) 0.000 011 848 556 746 899 456 × 2 = 0 + 0.000 023 697 113 493 798 912;
  • 28) 0.000 023 697 113 493 798 912 × 2 = 0 + 0.000 047 394 226 987 597 824;
  • 29) 0.000 047 394 226 987 597 824 × 2 = 0 + 0.000 094 788 453 975 195 648;
  • 30) 0.000 094 788 453 975 195 648 × 2 = 0 + 0.000 189 576 907 950 391 296;
  • 31) 0.000 189 576 907 950 391 296 × 2 = 0 + 0.000 379 153 815 900 782 592;
  • 32) 0.000 379 153 815 900 782 592 × 2 = 0 + 0.000 758 307 631 801 565 184;
  • 33) 0.000 758 307 631 801 565 184 × 2 = 0 + 0.001 516 615 263 603 130 368;
  • 34) 0.001 516 615 263 603 130 368 × 2 = 0 + 0.003 033 230 527 206 260 736;
  • 35) 0.003 033 230 527 206 260 736 × 2 = 0 + 0.006 066 461 054 412 521 472;
  • 36) 0.006 066 461 054 412 521 472 × 2 = 0 + 0.012 132 922 108 825 042 944;
  • 37) 0.012 132 922 108 825 042 944 × 2 = 0 + 0.024 265 844 217 650 085 888;
  • 38) 0.024 265 844 217 650 085 888 × 2 = 0 + 0.048 531 688 435 300 171 776;
  • 39) 0.048 531 688 435 300 171 776 × 2 = 0 + 0.097 063 376 870 600 343 552;
  • 40) 0.097 063 376 870 600 343 552 × 2 = 0 + 0.194 126 753 741 200 687 104;
  • 41) 0.194 126 753 741 200 687 104 × 2 = 0 + 0.388 253 507 482 401 374 208;
  • 42) 0.388 253 507 482 401 374 208 × 2 = 0 + 0.776 507 014 964 802 748 416;
  • 43) 0.776 507 014 964 802 748 416 × 2 = 1 + 0.553 014 029 929 605 496 832;
  • 44) 0.553 014 029 929 605 496 832 × 2 = 1 + 0.106 028 059 859 210 993 664;
  • 45) 0.106 028 059 859 210 993 664 × 2 = 0 + 0.212 056 119 718 421 987 328;
  • 46) 0.212 056 119 718 421 987 328 × 2 = 0 + 0.424 112 239 436 843 974 656;
  • 47) 0.424 112 239 436 843 974 656 × 2 = 0 + 0.848 224 478 873 687 949 312;
  • 48) 0.848 224 478 873 687 949 312 × 2 = 1 + 0.696 448 957 747 375 898 624;
  • 49) 0.696 448 957 747 375 898 624 × 2 = 1 + 0.392 897 915 494 751 797 248;
  • 50) 0.392 897 915 494 751 797 248 × 2 = 0 + 0.785 795 830 989 503 594 496;
  • 51) 0.785 795 830 989 503 594 496 × 2 = 1 + 0.571 591 661 979 007 188 992;
  • 52) 0.571 591 661 979 007 188 992 × 2 = 1 + 0.143 183 323 958 014 377 984;
  • 53) 0.143 183 323 958 014 377 984 × 2 = 0 + 0.286 366 647 916 028 755 968;
  • 54) 0.286 366 647 916 028 755 968 × 2 = 0 + 0.572 733 295 832 057 511 936;
  • 55) 0.572 733 295 832 057 511 936 × 2 = 1 + 0.145 466 591 664 115 023 872;
  • 56) 0.145 466 591 664 115 023 872 × 2 = 0 + 0.290 933 183 328 230 047 744;
  • 57) 0.290 933 183 328 230 047 744 × 2 = 0 + 0.581 866 366 656 460 095 488;
  • 58) 0.581 866 366 656 460 095 488 × 2 = 1 + 0.163 732 733 312 920 190 976;
  • 59) 0.163 732 733 312 920 190 976 × 2 = 0 + 0.327 465 466 625 840 381 952;
  • 60) 0.327 465 466 625 840 381 952 × 2 = 0 + 0.654 930 933 251 680 763 904;
  • 61) 0.654 930 933 251 680 763 904 × 2 = 1 + 0.309 861 866 503 361 527 808;
  • 62) 0.309 861 866 503 361 527 808 × 2 = 0 + 0.619 723 733 006 723 055 616;
  • 63) 0.619 723 733 006 723 055 616 × 2 = 1 + 0.239 447 466 013 446 111 232;
  • 64) 0.239 447 466 013 446 111 232 × 2 = 0 + 0.478 894 932 026 892 222 464;
  • 65) 0.478 894 932 026 892 222 464 × 2 = 0 + 0.957 789 864 053 784 444 928;
  • 66) 0.957 789 864 053 784 444 928 × 2 = 1 + 0.915 579 728 107 568 889 856;
  • 67) 0.915 579 728 107 568 889 856 × 2 = 1 + 0.831 159 456 215 137 779 712;
  • 68) 0.831 159 456 215 137 779 712 × 2 = 1 + 0.662 318 912 430 275 559 424;
  • 69) 0.662 318 912 430 275 559 424 × 2 = 1 + 0.324 637 824 860 551 118 848;
  • 70) 0.324 637 824 860 551 118 848 × 2 = 0 + 0.649 275 649 721 102 237 696;
  • 71) 0.649 275 649 721 102 237 696 × 2 = 1 + 0.298 551 299 442 204 475 392;
  • 72) 0.298 551 299 442 204 475 392 × 2 = 0 + 0.597 102 598 884 408 950 784;
  • 73) 0.597 102 598 884 408 950 784 × 2 = 1 + 0.194 205 197 768 817 901 568;
  • 74) 0.194 205 197 768 817 901 568 × 2 = 0 + 0.388 410 395 537 635 803 136;
  • 75) 0.388 410 395 537 635 803 136 × 2 = 0 + 0.776 820 791 075 271 606 272;
  • 76) 0.776 820 791 075 271 606 272 × 2 = 1 + 0.553 641 582 150 543 212 544;
  • 77) 0.553 641 582 150 543 212 544 × 2 = 1 + 0.107 283 164 301 086 425 088;
  • 78) 0.107 283 164 301 086 425 088 × 2 = 0 + 0.214 566 328 602 172 850 176;
  • 79) 0.214 566 328 602 172 850 176 × 2 = 0 + 0.429 132 657 204 345 700 352;
  • 80) 0.429 132 657 204 345 700 352 × 2 = 0 + 0.858 265 314 408 691 400 704;
  • 81) 0.858 265 314 408 691 400 704 × 2 = 1 + 0.716 530 628 817 382 801 408;
  • 82) 0.716 530 628 817 382 801 408 × 2 = 1 + 0.433 061 257 634 765 602 816;
  • 83) 0.433 061 257 634 765 602 816 × 2 = 0 + 0.866 122 515 269 531 205 632;
  • 84) 0.866 122 515 269 531 205 632 × 2 = 1 + 0.732 245 030 539 062 411 264;
  • 85) 0.732 245 030 539 062 411 264 × 2 = 1 + 0.464 490 061 078 124 822 528;
  • 86) 0.464 490 061 078 124 822 528 × 2 = 0 + 0.928 980 122 156 249 645 056;
  • 87) 0.928 980 122 156 249 645 056 × 2 = 1 + 0.857 960 244 312 499 290 112;
  • 88) 0.857 960 244 312 499 290 112 × 2 = 1 + 0.715 920 488 624 998 580 224;
  • 89) 0.715 920 488 624 998 580 224 × 2 = 1 + 0.431 840 977 249 997 160 448;
  • 90) 0.431 840 977 249 997 160 448 × 2 = 0 + 0.863 681 954 499 994 320 896;
  • 91) 0.863 681 954 499 994 320 896 × 2 = 1 + 0.727 363 908 999 988 641 792;
  • 92) 0.727 363 908 999 988 641 792 × 2 = 1 + 0.454 727 817 999 977 283 584;
  • 93) 0.454 727 817 999 977 283 584 × 2 = 0 + 0.909 455 635 999 954 567 168;
  • 94) 0.909 455 635 999 954 567 168 × 2 = 1 + 0.818 911 271 999 909 134 336;
  • 95) 0.818 911 271 999 909 134 336 × 2 = 1 + 0.637 822 543 999 818 268 672;

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:


0.000 000 000 000 176 557 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0111 1010 1001 1000 1101 1011 1011 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0111 1010 1001 1000 1101 1011 1011 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:


0.000 000 000 000 176 557 254(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0111 1010 1001 1000 1101 1011 1011 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1010 0111 1010 1001 1000 1101 1011 1011 011(2) × 20 =

1.1000 1101 1001 0010 0101 0011 1101 0100 1100 0110 1101 1101 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 0101 0011 1101 0100 1100 0110 1101 1101 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 0101 0011 1101 0100 1100 0110 1101 1101 1011 =

1000 1101 1001 0010 0101 0011 1101 0100 1100 0110 1101 1101 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 0101 0011 1101 0100 1100 0110 1101 1101 1011


Decimal number -0.000 000 000 000 176 557 254 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 0011 1101 0100 1100 0110 1101 1101 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