-0.000 000 000 000 176 557 331 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 331(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
-0.000 000 000 000 176 557 331(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:

|-0.000 000 000 000 176 557 331| = 0.000 000 000 000 176 557 331


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: 0.000 000 000 000 176 557 331.

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) 0.000 000 000 000 176 557 331 × 2 = 0 + 0.000 000 000 000 353 114 662;
  • 2) 0.000 000 000 000 353 114 662 × 2 = 0 + 0.000 000 000 000 706 229 324;
  • 3) 0.000 000 000 000 706 229 324 × 2 = 0 + 0.000 000 000 001 412 458 648;
  • 4) 0.000 000 000 001 412 458 648 × 2 = 0 + 0.000 000 000 002 824 917 296;
  • 5) 0.000 000 000 002 824 917 296 × 2 = 0 + 0.000 000 000 005 649 834 592;
  • 6) 0.000 000 000 005 649 834 592 × 2 = 0 + 0.000 000 000 011 299 669 184;
  • 7) 0.000 000 000 011 299 669 184 × 2 = 0 + 0.000 000 000 022 599 338 368;
  • 8) 0.000 000 000 022 599 338 368 × 2 = 0 + 0.000 000 000 045 198 676 736;
  • 9) 0.000 000 000 045 198 676 736 × 2 = 0 + 0.000 000 000 090 397 353 472;
  • 10) 0.000 000 000 090 397 353 472 × 2 = 0 + 0.000 000 000 180 794 706 944;
  • 11) 0.000 000 000 180 794 706 944 × 2 = 0 + 0.000 000 000 361 589 413 888;
  • 12) 0.000 000 000 361 589 413 888 × 2 = 0 + 0.000 000 000 723 178 827 776;
  • 13) 0.000 000 000 723 178 827 776 × 2 = 0 + 0.000 000 001 446 357 655 552;
  • 14) 0.000 000 001 446 357 655 552 × 2 = 0 + 0.000 000 002 892 715 311 104;
  • 15) 0.000 000 002 892 715 311 104 × 2 = 0 + 0.000 000 005 785 430 622 208;
  • 16) 0.000 000 005 785 430 622 208 × 2 = 0 + 0.000 000 011 570 861 244 416;
  • 17) 0.000 000 011 570 861 244 416 × 2 = 0 + 0.000 000 023 141 722 488 832;
  • 18) 0.000 000 023 141 722 488 832 × 2 = 0 + 0.000 000 046 283 444 977 664;
  • 19) 0.000 000 046 283 444 977 664 × 2 = 0 + 0.000 000 092 566 889 955 328;
  • 20) 0.000 000 092 566 889 955 328 × 2 = 0 + 0.000 000 185 133 779 910 656;
  • 21) 0.000 000 185 133 779 910 656 × 2 = 0 + 0.000 000 370 267 559 821 312;
  • 22) 0.000 000 370 267 559 821 312 × 2 = 0 + 0.000 000 740 535 119 642 624;
  • 23) 0.000 000 740 535 119 642 624 × 2 = 0 + 0.000 001 481 070 239 285 248;
  • 24) 0.000 001 481 070 239 285 248 × 2 = 0 + 0.000 002 962 140 478 570 496;
  • 25) 0.000 002 962 140 478 570 496 × 2 = 0 + 0.000 005 924 280 957 140 992;
  • 26) 0.000 005 924 280 957 140 992 × 2 = 0 + 0.000 011 848 561 914 281 984;
  • 27) 0.000 011 848 561 914 281 984 × 2 = 0 + 0.000 023 697 123 828 563 968;
  • 28) 0.000 023 697 123 828 563 968 × 2 = 0 + 0.000 047 394 247 657 127 936;
  • 29) 0.000 047 394 247 657 127 936 × 2 = 0 + 0.000 094 788 495 314 255 872;
  • 30) 0.000 094 788 495 314 255 872 × 2 = 0 + 0.000 189 576 990 628 511 744;
  • 31) 0.000 189 576 990 628 511 744 × 2 = 0 + 0.000 379 153 981 257 023 488;
  • 32) 0.000 379 153 981 257 023 488 × 2 = 0 + 0.000 758 307 962 514 046 976;
  • 33) 0.000 758 307 962 514 046 976 × 2 = 0 + 0.001 516 615 925 028 093 952;
  • 34) 0.001 516 615 925 028 093 952 × 2 = 0 + 0.003 033 231 850 056 187 904;
  • 35) 0.003 033 231 850 056 187 904 × 2 = 0 + 0.006 066 463 700 112 375 808;
  • 36) 0.006 066 463 700 112 375 808 × 2 = 0 + 0.012 132 927 400 224 751 616;
  • 37) 0.012 132 927 400 224 751 616 × 2 = 0 + 0.024 265 854 800 449 503 232;
  • 38) 0.024 265 854 800 449 503 232 × 2 = 0 + 0.048 531 709 600 899 006 464;
  • 39) 0.048 531 709 600 899 006 464 × 2 = 0 + 0.097 063 419 201 798 012 928;
  • 40) 0.097 063 419 201 798 012 928 × 2 = 0 + 0.194 126 838 403 596 025 856;
  • 41) 0.194 126 838 403 596 025 856 × 2 = 0 + 0.388 253 676 807 192 051 712;
  • 42) 0.388 253 676 807 192 051 712 × 2 = 0 + 0.776 507 353 614 384 103 424;
  • 43) 0.776 507 353 614 384 103 424 × 2 = 1 + 0.553 014 707 228 768 206 848;
  • 44) 0.553 014 707 228 768 206 848 × 2 = 1 + 0.106 029 414 457 536 413 696;
  • 45) 0.106 029 414 457 536 413 696 × 2 = 0 + 0.212 058 828 915 072 827 392;
  • 46) 0.212 058 828 915 072 827 392 × 2 = 0 + 0.424 117 657 830 145 654 784;
  • 47) 0.424 117 657 830 145 654 784 × 2 = 0 + 0.848 235 315 660 291 309 568;
  • 48) 0.848 235 315 660 291 309 568 × 2 = 1 + 0.696 470 631 320 582 619 136;
  • 49) 0.696 470 631 320 582 619 136 × 2 = 1 + 0.392 941 262 641 165 238 272;
  • 50) 0.392 941 262 641 165 238 272 × 2 = 0 + 0.785 882 525 282 330 476 544;
  • 51) 0.785 882 525 282 330 476 544 × 2 = 1 + 0.571 765 050 564 660 953 088;
  • 52) 0.571 765 050 564 660 953 088 × 2 = 1 + 0.143 530 101 129 321 906 176;
  • 53) 0.143 530 101 129 321 906 176 × 2 = 0 + 0.287 060 202 258 643 812 352;
  • 54) 0.287 060 202 258 643 812 352 × 2 = 0 + 0.574 120 404 517 287 624 704;
  • 55) 0.574 120 404 517 287 624 704 × 2 = 1 + 0.148 240 809 034 575 249 408;
  • 56) 0.148 240 809 034 575 249 408 × 2 = 0 + 0.296 481 618 069 150 498 816;
  • 57) 0.296 481 618 069 150 498 816 × 2 = 0 + 0.592 963 236 138 300 997 632;
  • 58) 0.592 963 236 138 300 997 632 × 2 = 1 + 0.185 926 472 276 601 995 264;
  • 59) 0.185 926 472 276 601 995 264 × 2 = 0 + 0.371 852 944 553 203 990 528;
  • 60) 0.371 852 944 553 203 990 528 × 2 = 0 + 0.743 705 889 106 407 981 056;
  • 61) 0.743 705 889 106 407 981 056 × 2 = 1 + 0.487 411 778 212 815 962 112;
  • 62) 0.487 411 778 212 815 962 112 × 2 = 0 + 0.974 823 556 425 631 924 224;
  • 63) 0.974 823 556 425 631 924 224 × 2 = 1 + 0.949 647 112 851 263 848 448;
  • 64) 0.949 647 112 851 263 848 448 × 2 = 1 + 0.899 294 225 702 527 696 896;
  • 65) 0.899 294 225 702 527 696 896 × 2 = 1 + 0.798 588 451 405 055 393 792;
  • 66) 0.798 588 451 405 055 393 792 × 2 = 1 + 0.597 176 902 810 110 787 584;
  • 67) 0.597 176 902 810 110 787 584 × 2 = 1 + 0.194 353 805 620 221 575 168;
  • 68) 0.194 353 805 620 221 575 168 × 2 = 0 + 0.388 707 611 240 443 150 336;
  • 69) 0.388 707 611 240 443 150 336 × 2 = 0 + 0.777 415 222 480 886 300 672;
  • 70) 0.777 415 222 480 886 300 672 × 2 = 1 + 0.554 830 444 961 772 601 344;
  • 71) 0.554 830 444 961 772 601 344 × 2 = 1 + 0.109 660 889 923 545 202 688;
  • 72) 0.109 660 889 923 545 202 688 × 2 = 0 + 0.219 321 779 847 090 405 376;
  • 73) 0.219 321 779 847 090 405 376 × 2 = 0 + 0.438 643 559 694 180 810 752;
  • 74) 0.438 643 559 694 180 810 752 × 2 = 0 + 0.877 287 119 388 361 621 504;
  • 75) 0.877 287 119 388 361 621 504 × 2 = 1 + 0.754 574 238 776 723 243 008;
  • 76) 0.754 574 238 776 723 243 008 × 2 = 1 + 0.509 148 477 553 446 486 016;
  • 77) 0.509 148 477 553 446 486 016 × 2 = 1 + 0.018 296 955 106 892 972 032;
  • 78) 0.018 296 955 106 892 972 032 × 2 = 0 + 0.036 593 910 213 785 944 064;
  • 79) 0.036 593 910 213 785 944 064 × 2 = 0 + 0.073 187 820 427 571 888 128;
  • 80) 0.073 187 820 427 571 888 128 × 2 = 0 + 0.146 375 640 855 143 776 256;
  • 81) 0.146 375 640 855 143 776 256 × 2 = 0 + 0.292 751 281 710 287 552 512;
  • 82) 0.292 751 281 710 287 552 512 × 2 = 0 + 0.585 502 563 420 575 105 024;
  • 83) 0.585 502 563 420 575 105 024 × 2 = 1 + 0.171 005 126 841 150 210 048;
  • 84) 0.171 005 126 841 150 210 048 × 2 = 0 + 0.342 010 253 682 300 420 096;
  • 85) 0.342 010 253 682 300 420 096 × 2 = 0 + 0.684 020 507 364 600 840 192;
  • 86) 0.684 020 507 364 600 840 192 × 2 = 1 + 0.368 041 014 729 201 680 384;
  • 87) 0.368 041 014 729 201 680 384 × 2 = 0 + 0.736 082 029 458 403 360 768;
  • 88) 0.736 082 029 458 403 360 768 × 2 = 1 + 0.472 164 058 916 806 721 536;
  • 89) 0.472 164 058 916 806 721 536 × 2 = 0 + 0.944 328 117 833 613 443 072;
  • 90) 0.944 328 117 833 613 443 072 × 2 = 1 + 0.888 656 235 667 226 886 144;
  • 91) 0.888 656 235 667 226 886 144 × 2 = 1 + 0.777 312 471 334 453 772 288;
  • 92) 0.777 312 471 334 453 772 288 × 2 = 1 + 0.554 624 942 668 907 544 576;
  • 93) 0.554 624 942 668 907 544 576 × 2 = 1 + 0.109 249 885 337 815 089 152;
  • 94) 0.109 249 885 337 815 089 152 × 2 = 0 + 0.218 499 770 675 630 178 304;
  • 95) 0.218 499 770 675 630 178 304 × 2 = 0 + 0.436 999 541 351 260 356 608;

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:


0.000 000 000 000 176 557 331(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1110 0110 0011 1000 0010 0101 0111 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 331(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1110 0110 0011 1000 0010 0101 0111 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:


0.000 000 000 000 176 557 331(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1110 0110 0011 1000 0010 0101 0111 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1011 1110 0110 0011 1000 0010 0101 0111 100(2) × 20 =

1.1000 1101 1001 0010 0101 1111 0011 0001 1100 0001 0010 1011 1100(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 1111 0011 0001 1100 0001 0010 1011 1100


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 1111 0011 0001 1100 0001 0010 1011 1100 =

1000 1101 1001 0010 0101 1111 0011 0001 1100 0001 0010 1011 1100


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 1111 0011 0001 1100 0001 0010 1011 1100


Decimal number -0.000 000 000 000 176 557 331 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0101 1111 0011 0001 1100 0001 0010 1011 1100

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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