-0.000 000 000 000 176 557 196 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 196(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
-0.000 000 000 000 176 557 196(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:

|-0.000 000 000 000 176 557 196| = 0.000 000 000 000 176 557 196


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: 0.000 000 000 000 176 557 196.

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) 0.000 000 000 000 176 557 196 × 2 = 0 + 0.000 000 000 000 353 114 392;
  • 2) 0.000 000 000 000 353 114 392 × 2 = 0 + 0.000 000 000 000 706 228 784;
  • 3) 0.000 000 000 000 706 228 784 × 2 = 0 + 0.000 000 000 001 412 457 568;
  • 4) 0.000 000 000 001 412 457 568 × 2 = 0 + 0.000 000 000 002 824 915 136;
  • 5) 0.000 000 000 002 824 915 136 × 2 = 0 + 0.000 000 000 005 649 830 272;
  • 6) 0.000 000 000 005 649 830 272 × 2 = 0 + 0.000 000 000 011 299 660 544;
  • 7) 0.000 000 000 011 299 660 544 × 2 = 0 + 0.000 000 000 022 599 321 088;
  • 8) 0.000 000 000 022 599 321 088 × 2 = 0 + 0.000 000 000 045 198 642 176;
  • 9) 0.000 000 000 045 198 642 176 × 2 = 0 + 0.000 000 000 090 397 284 352;
  • 10) 0.000 000 000 090 397 284 352 × 2 = 0 + 0.000 000 000 180 794 568 704;
  • 11) 0.000 000 000 180 794 568 704 × 2 = 0 + 0.000 000 000 361 589 137 408;
  • 12) 0.000 000 000 361 589 137 408 × 2 = 0 + 0.000 000 000 723 178 274 816;
  • 13) 0.000 000 000 723 178 274 816 × 2 = 0 + 0.000 000 001 446 356 549 632;
  • 14) 0.000 000 001 446 356 549 632 × 2 = 0 + 0.000 000 002 892 713 099 264;
  • 15) 0.000 000 002 892 713 099 264 × 2 = 0 + 0.000 000 005 785 426 198 528;
  • 16) 0.000 000 005 785 426 198 528 × 2 = 0 + 0.000 000 011 570 852 397 056;
  • 17) 0.000 000 011 570 852 397 056 × 2 = 0 + 0.000 000 023 141 704 794 112;
  • 18) 0.000 000 023 141 704 794 112 × 2 = 0 + 0.000 000 046 283 409 588 224;
  • 19) 0.000 000 046 283 409 588 224 × 2 = 0 + 0.000 000 092 566 819 176 448;
  • 20) 0.000 000 092 566 819 176 448 × 2 = 0 + 0.000 000 185 133 638 352 896;
  • 21) 0.000 000 185 133 638 352 896 × 2 = 0 + 0.000 000 370 267 276 705 792;
  • 22) 0.000 000 370 267 276 705 792 × 2 = 0 + 0.000 000 740 534 553 411 584;
  • 23) 0.000 000 740 534 553 411 584 × 2 = 0 + 0.000 001 481 069 106 823 168;
  • 24) 0.000 001 481 069 106 823 168 × 2 = 0 + 0.000 002 962 138 213 646 336;
  • 25) 0.000 002 962 138 213 646 336 × 2 = 0 + 0.000 005 924 276 427 292 672;
  • 26) 0.000 005 924 276 427 292 672 × 2 = 0 + 0.000 011 848 552 854 585 344;
  • 27) 0.000 011 848 552 854 585 344 × 2 = 0 + 0.000 023 697 105 709 170 688;
  • 28) 0.000 023 697 105 709 170 688 × 2 = 0 + 0.000 047 394 211 418 341 376;
  • 29) 0.000 047 394 211 418 341 376 × 2 = 0 + 0.000 094 788 422 836 682 752;
  • 30) 0.000 094 788 422 836 682 752 × 2 = 0 + 0.000 189 576 845 673 365 504;
  • 31) 0.000 189 576 845 673 365 504 × 2 = 0 + 0.000 379 153 691 346 731 008;
  • 32) 0.000 379 153 691 346 731 008 × 2 = 0 + 0.000 758 307 382 693 462 016;
  • 33) 0.000 758 307 382 693 462 016 × 2 = 0 + 0.001 516 614 765 386 924 032;
  • 34) 0.001 516 614 765 386 924 032 × 2 = 0 + 0.003 033 229 530 773 848 064;
  • 35) 0.003 033 229 530 773 848 064 × 2 = 0 + 0.006 066 459 061 547 696 128;
  • 36) 0.006 066 459 061 547 696 128 × 2 = 0 + 0.012 132 918 123 095 392 256;
  • 37) 0.012 132 918 123 095 392 256 × 2 = 0 + 0.024 265 836 246 190 784 512;
  • 38) 0.024 265 836 246 190 784 512 × 2 = 0 + 0.048 531 672 492 381 569 024;
  • 39) 0.048 531 672 492 381 569 024 × 2 = 0 + 0.097 063 344 984 763 138 048;
  • 40) 0.097 063 344 984 763 138 048 × 2 = 0 + 0.194 126 689 969 526 276 096;
  • 41) 0.194 126 689 969 526 276 096 × 2 = 0 + 0.388 253 379 939 052 552 192;
  • 42) 0.388 253 379 939 052 552 192 × 2 = 0 + 0.776 506 759 878 105 104 384;
  • 43) 0.776 506 759 878 105 104 384 × 2 = 1 + 0.553 013 519 756 210 208 768;
  • 44) 0.553 013 519 756 210 208 768 × 2 = 1 + 0.106 027 039 512 420 417 536;
  • 45) 0.106 027 039 512 420 417 536 × 2 = 0 + 0.212 054 079 024 840 835 072;
  • 46) 0.212 054 079 024 840 835 072 × 2 = 0 + 0.424 108 158 049 681 670 144;
  • 47) 0.424 108 158 049 681 670 144 × 2 = 0 + 0.848 216 316 099 363 340 288;
  • 48) 0.848 216 316 099 363 340 288 × 2 = 1 + 0.696 432 632 198 726 680 576;
  • 49) 0.696 432 632 198 726 680 576 × 2 = 1 + 0.392 865 264 397 453 361 152;
  • 50) 0.392 865 264 397 453 361 152 × 2 = 0 + 0.785 730 528 794 906 722 304;
  • 51) 0.785 730 528 794 906 722 304 × 2 = 1 + 0.571 461 057 589 813 444 608;
  • 52) 0.571 461 057 589 813 444 608 × 2 = 1 + 0.142 922 115 179 626 889 216;
  • 53) 0.142 922 115 179 626 889 216 × 2 = 0 + 0.285 844 230 359 253 778 432;
  • 54) 0.285 844 230 359 253 778 432 × 2 = 0 + 0.571 688 460 718 507 556 864;
  • 55) 0.571 688 460 718 507 556 864 × 2 = 1 + 0.143 376 921 437 015 113 728;
  • 56) 0.143 376 921 437 015 113 728 × 2 = 0 + 0.286 753 842 874 030 227 456;
  • 57) 0.286 753 842 874 030 227 456 × 2 = 0 + 0.573 507 685 748 060 454 912;
  • 58) 0.573 507 685 748 060 454 912 × 2 = 1 + 0.147 015 371 496 120 909 824;
  • 59) 0.147 015 371 496 120 909 824 × 2 = 0 + 0.294 030 742 992 241 819 648;
  • 60) 0.294 030 742 992 241 819 648 × 2 = 0 + 0.588 061 485 984 483 639 296;
  • 61) 0.588 061 485 984 483 639 296 × 2 = 1 + 0.176 122 971 968 967 278 592;
  • 62) 0.176 122 971 968 967 278 592 × 2 = 0 + 0.352 245 943 937 934 557 184;
  • 63) 0.352 245 943 937 934 557 184 × 2 = 0 + 0.704 491 887 875 869 114 368;
  • 64) 0.704 491 887 875 869 114 368 × 2 = 1 + 0.408 983 775 751 738 228 736;
  • 65) 0.408 983 775 751 738 228 736 × 2 = 0 + 0.817 967 551 503 476 457 472;
  • 66) 0.817 967 551 503 476 457 472 × 2 = 1 + 0.635 935 103 006 952 914 944;
  • 67) 0.635 935 103 006 952 914 944 × 2 = 1 + 0.271 870 206 013 905 829 888;
  • 68) 0.271 870 206 013 905 829 888 × 2 = 0 + 0.543 740 412 027 811 659 776;
  • 69) 0.543 740 412 027 811 659 776 × 2 = 1 + 0.087 480 824 055 623 319 552;
  • 70) 0.087 480 824 055 623 319 552 × 2 = 0 + 0.174 961 648 111 246 639 104;
  • 71) 0.174 961 648 111 246 639 104 × 2 = 0 + 0.349 923 296 222 493 278 208;
  • 72) 0.349 923 296 222 493 278 208 × 2 = 0 + 0.699 846 592 444 986 556 416;
  • 73) 0.699 846 592 444 986 556 416 × 2 = 1 + 0.399 693 184 889 973 112 832;
  • 74) 0.399 693 184 889 973 112 832 × 2 = 0 + 0.799 386 369 779 946 225 664;
  • 75) 0.799 386 369 779 946 225 664 × 2 = 1 + 0.598 772 739 559 892 451 328;
  • 76) 0.598 772 739 559 892 451 328 × 2 = 1 + 0.197 545 479 119 784 902 656;
  • 77) 0.197 545 479 119 784 902 656 × 2 = 0 + 0.395 090 958 239 569 805 312;
  • 78) 0.395 090 958 239 569 805 312 × 2 = 0 + 0.790 181 916 479 139 610 624;
  • 79) 0.790 181 916 479 139 610 624 × 2 = 1 + 0.580 363 832 958 279 221 248;
  • 80) 0.580 363 832 958 279 221 248 × 2 = 1 + 0.160 727 665 916 558 442 496;
  • 81) 0.160 727 665 916 558 442 496 × 2 = 0 + 0.321 455 331 833 116 884 992;
  • 82) 0.321 455 331 833 116 884 992 × 2 = 0 + 0.642 910 663 666 233 769 984;
  • 83) 0.642 910 663 666 233 769 984 × 2 = 1 + 0.285 821 327 332 467 539 968;
  • 84) 0.285 821 327 332 467 539 968 × 2 = 0 + 0.571 642 654 664 935 079 936;
  • 85) 0.571 642 654 664 935 079 936 × 2 = 1 + 0.143 285 309 329 870 159 872;
  • 86) 0.143 285 309 329 870 159 872 × 2 = 0 + 0.286 570 618 659 740 319 744;
  • 87) 0.286 570 618 659 740 319 744 × 2 = 0 + 0.573 141 237 319 480 639 488;
  • 88) 0.573 141 237 319 480 639 488 × 2 = 1 + 0.146 282 474 638 961 278 976;
  • 89) 0.146 282 474 638 961 278 976 × 2 = 0 + 0.292 564 949 277 922 557 952;
  • 90) 0.292 564 949 277 922 557 952 × 2 = 0 + 0.585 129 898 555 845 115 904;
  • 91) 0.585 129 898 555 845 115 904 × 2 = 1 + 0.170 259 797 111 690 231 808;
  • 92) 0.170 259 797 111 690 231 808 × 2 = 0 + 0.340 519 594 223 380 463 616;
  • 93) 0.340 519 594 223 380 463 616 × 2 = 0 + 0.681 039 188 446 760 927 232;
  • 94) 0.681 039 188 446 760 927 232 × 2 = 1 + 0.362 078 376 893 521 854 464;
  • 95) 0.362 078 376 893 521 854 464 × 2 = 0 + 0.724 156 753 787 043 708 928;

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:


0.000 000 000 000 176 557 196(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0110 1000 1011 0011 0010 1001 0010 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 196(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0110 1000 1011 0011 0010 1001 0010 010(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:


0.000 000 000 000 176 557 196(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0110 1000 1011 0011 0010 1001 0010 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0110 1000 1011 0011 0010 1001 0010 010(2) × 20 =

1.1000 1101 1001 0010 0100 1011 0100 0101 1001 1001 0100 1001 0010(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 0100 1011 0100 0101 1001 1001 0100 1001 0010


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 0100 1011 0100 0101 1001 1001 0100 1001 0010 =

1000 1101 1001 0010 0100 1011 0100 0101 1001 1001 0100 1001 0010


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 0100 1011 0100 0101 1001 1001 0100 1001 0010


Decimal number -0.000 000 000 000 176 557 196 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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