-0.000 000 000 000 176 546 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 546 9(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
-0.000 000 000 000 176 546 9(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:

|-0.000 000 000 000 176 546 9| = 0.000 000 000 000 176 546 9


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: 0.000 000 000 000 176 546 9.

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) 0.000 000 000 000 176 546 9 × 2 = 0 + 0.000 000 000 000 353 093 8;
  • 2) 0.000 000 000 000 353 093 8 × 2 = 0 + 0.000 000 000 000 706 187 6;
  • 3) 0.000 000 000 000 706 187 6 × 2 = 0 + 0.000 000 000 001 412 375 2;
  • 4) 0.000 000 000 001 412 375 2 × 2 = 0 + 0.000 000 000 002 824 750 4;
  • 5) 0.000 000 000 002 824 750 4 × 2 = 0 + 0.000 000 000 005 649 500 8;
  • 6) 0.000 000 000 005 649 500 8 × 2 = 0 + 0.000 000 000 011 299 001 6;
  • 7) 0.000 000 000 011 299 001 6 × 2 = 0 + 0.000 000 000 022 598 003 2;
  • 8) 0.000 000 000 022 598 003 2 × 2 = 0 + 0.000 000 000 045 196 006 4;
  • 9) 0.000 000 000 045 196 006 4 × 2 = 0 + 0.000 000 000 090 392 012 8;
  • 10) 0.000 000 000 090 392 012 8 × 2 = 0 + 0.000 000 000 180 784 025 6;
  • 11) 0.000 000 000 180 784 025 6 × 2 = 0 + 0.000 000 000 361 568 051 2;
  • 12) 0.000 000 000 361 568 051 2 × 2 = 0 + 0.000 000 000 723 136 102 4;
  • 13) 0.000 000 000 723 136 102 4 × 2 = 0 + 0.000 000 001 446 272 204 8;
  • 14) 0.000 000 001 446 272 204 8 × 2 = 0 + 0.000 000 002 892 544 409 6;
  • 15) 0.000 000 002 892 544 409 6 × 2 = 0 + 0.000 000 005 785 088 819 2;
  • 16) 0.000 000 005 785 088 819 2 × 2 = 0 + 0.000 000 011 570 177 638 4;
  • 17) 0.000 000 011 570 177 638 4 × 2 = 0 + 0.000 000 023 140 355 276 8;
  • 18) 0.000 000 023 140 355 276 8 × 2 = 0 + 0.000 000 046 280 710 553 6;
  • 19) 0.000 000 046 280 710 553 6 × 2 = 0 + 0.000 000 092 561 421 107 2;
  • 20) 0.000 000 092 561 421 107 2 × 2 = 0 + 0.000 000 185 122 842 214 4;
  • 21) 0.000 000 185 122 842 214 4 × 2 = 0 + 0.000 000 370 245 684 428 8;
  • 22) 0.000 000 370 245 684 428 8 × 2 = 0 + 0.000 000 740 491 368 857 6;
  • 23) 0.000 000 740 491 368 857 6 × 2 = 0 + 0.000 001 480 982 737 715 2;
  • 24) 0.000 001 480 982 737 715 2 × 2 = 0 + 0.000 002 961 965 475 430 4;
  • 25) 0.000 002 961 965 475 430 4 × 2 = 0 + 0.000 005 923 930 950 860 8;
  • 26) 0.000 005 923 930 950 860 8 × 2 = 0 + 0.000 011 847 861 901 721 6;
  • 27) 0.000 011 847 861 901 721 6 × 2 = 0 + 0.000 023 695 723 803 443 2;
  • 28) 0.000 023 695 723 803 443 2 × 2 = 0 + 0.000 047 391 447 606 886 4;
  • 29) 0.000 047 391 447 606 886 4 × 2 = 0 + 0.000 094 782 895 213 772 8;
  • 30) 0.000 094 782 895 213 772 8 × 2 = 0 + 0.000 189 565 790 427 545 6;
  • 31) 0.000 189 565 790 427 545 6 × 2 = 0 + 0.000 379 131 580 855 091 2;
  • 32) 0.000 379 131 580 855 091 2 × 2 = 0 + 0.000 758 263 161 710 182 4;
  • 33) 0.000 758 263 161 710 182 4 × 2 = 0 + 0.001 516 526 323 420 364 8;
  • 34) 0.001 516 526 323 420 364 8 × 2 = 0 + 0.003 033 052 646 840 729 6;
  • 35) 0.003 033 052 646 840 729 6 × 2 = 0 + 0.006 066 105 293 681 459 2;
  • 36) 0.006 066 105 293 681 459 2 × 2 = 0 + 0.012 132 210 587 362 918 4;
  • 37) 0.012 132 210 587 362 918 4 × 2 = 0 + 0.024 264 421 174 725 836 8;
  • 38) 0.024 264 421 174 725 836 8 × 2 = 0 + 0.048 528 842 349 451 673 6;
  • 39) 0.048 528 842 349 451 673 6 × 2 = 0 + 0.097 057 684 698 903 347 2;
  • 40) 0.097 057 684 698 903 347 2 × 2 = 0 + 0.194 115 369 397 806 694 4;
  • 41) 0.194 115 369 397 806 694 4 × 2 = 0 + 0.388 230 738 795 613 388 8;
  • 42) 0.388 230 738 795 613 388 8 × 2 = 0 + 0.776 461 477 591 226 777 6;
  • 43) 0.776 461 477 591 226 777 6 × 2 = 1 + 0.552 922 955 182 453 555 2;
  • 44) 0.552 922 955 182 453 555 2 × 2 = 1 + 0.105 845 910 364 907 110 4;
  • 45) 0.105 845 910 364 907 110 4 × 2 = 0 + 0.211 691 820 729 814 220 8;
  • 46) 0.211 691 820 729 814 220 8 × 2 = 0 + 0.423 383 641 459 628 441 6;
  • 47) 0.423 383 641 459 628 441 6 × 2 = 0 + 0.846 767 282 919 256 883 2;
  • 48) 0.846 767 282 919 256 883 2 × 2 = 1 + 0.693 534 565 838 513 766 4;
  • 49) 0.693 534 565 838 513 766 4 × 2 = 1 + 0.387 069 131 677 027 532 8;
  • 50) 0.387 069 131 677 027 532 8 × 2 = 0 + 0.774 138 263 354 055 065 6;
  • 51) 0.774 138 263 354 055 065 6 × 2 = 1 + 0.548 276 526 708 110 131 2;
  • 52) 0.548 276 526 708 110 131 2 × 2 = 1 + 0.096 553 053 416 220 262 4;
  • 53) 0.096 553 053 416 220 262 4 × 2 = 0 + 0.193 106 106 832 440 524 8;
  • 54) 0.193 106 106 832 440 524 8 × 2 = 0 + 0.386 212 213 664 881 049 6;
  • 55) 0.386 212 213 664 881 049 6 × 2 = 0 + 0.772 424 427 329 762 099 2;
  • 56) 0.772 424 427 329 762 099 2 × 2 = 1 + 0.544 848 854 659 524 198 4;
  • 57) 0.544 848 854 659 524 198 4 × 2 = 1 + 0.089 697 709 319 048 396 8;
  • 58) 0.089 697 709 319 048 396 8 × 2 = 0 + 0.179 395 418 638 096 793 6;
  • 59) 0.179 395 418 638 096 793 6 × 2 = 0 + 0.358 790 837 276 193 587 2;
  • 60) 0.358 790 837 276 193 587 2 × 2 = 0 + 0.717 581 674 552 387 174 4;
  • 61) 0.717 581 674 552 387 174 4 × 2 = 1 + 0.435 163 349 104 774 348 8;
  • 62) 0.435 163 349 104 774 348 8 × 2 = 0 + 0.870 326 698 209 548 697 6;
  • 63) 0.870 326 698 209 548 697 6 × 2 = 1 + 0.740 653 396 419 097 395 2;
  • 64) 0.740 653 396 419 097 395 2 × 2 = 1 + 0.481 306 792 838 194 790 4;
  • 65) 0.481 306 792 838 194 790 4 × 2 = 0 + 0.962 613 585 676 389 580 8;
  • 66) 0.962 613 585 676 389 580 8 × 2 = 1 + 0.925 227 171 352 779 161 6;
  • 67) 0.925 227 171 352 779 161 6 × 2 = 1 + 0.850 454 342 705 558 323 2;
  • 68) 0.850 454 342 705 558 323 2 × 2 = 1 + 0.700 908 685 411 116 646 4;
  • 69) 0.700 908 685 411 116 646 4 × 2 = 1 + 0.401 817 370 822 233 292 8;
  • 70) 0.401 817 370 822 233 292 8 × 2 = 0 + 0.803 634 741 644 466 585 6;
  • 71) 0.803 634 741 644 466 585 6 × 2 = 1 + 0.607 269 483 288 933 171 2;
  • 72) 0.607 269 483 288 933 171 2 × 2 = 1 + 0.214 538 966 577 866 342 4;
  • 73) 0.214 538 966 577 866 342 4 × 2 = 0 + 0.429 077 933 155 732 684 8;
  • 74) 0.429 077 933 155 732 684 8 × 2 = 0 + 0.858 155 866 311 465 369 6;
  • 75) 0.858 155 866 311 465 369 6 × 2 = 1 + 0.716 311 732 622 930 739 2;
  • 76) 0.716 311 732 622 930 739 2 × 2 = 1 + 0.432 623 465 245 861 478 4;
  • 77) 0.432 623 465 245 861 478 4 × 2 = 0 + 0.865 246 930 491 722 956 8;
  • 78) 0.865 246 930 491 722 956 8 × 2 = 1 + 0.730 493 860 983 445 913 6;
  • 79) 0.730 493 860 983 445 913 6 × 2 = 1 + 0.460 987 721 966 891 827 2;
  • 80) 0.460 987 721 966 891 827 2 × 2 = 0 + 0.921 975 443 933 783 654 4;
  • 81) 0.921 975 443 933 783 654 4 × 2 = 1 + 0.843 950 887 867 567 308 8;
  • 82) 0.843 950 887 867 567 308 8 × 2 = 1 + 0.687 901 775 735 134 617 6;
  • 83) 0.687 901 775 735 134 617 6 × 2 = 1 + 0.375 803 551 470 269 235 2;
  • 84) 0.375 803 551 470 269 235 2 × 2 = 0 + 0.751 607 102 940 538 470 4;
  • 85) 0.751 607 102 940 538 470 4 × 2 = 1 + 0.503 214 205 881 076 940 8;
  • 86) 0.503 214 205 881 076 940 8 × 2 = 1 + 0.006 428 411 762 153 881 6;
  • 87) 0.006 428 411 762 153 881 6 × 2 = 0 + 0.012 856 823 524 307 763 2;
  • 88) 0.012 856 823 524 307 763 2 × 2 = 0 + 0.025 713 647 048 615 526 4;
  • 89) 0.025 713 647 048 615 526 4 × 2 = 0 + 0.051 427 294 097 231 052 8;
  • 90) 0.051 427 294 097 231 052 8 × 2 = 0 + 0.102 854 588 194 462 105 6;
  • 91) 0.102 854 588 194 462 105 6 × 2 = 0 + 0.205 709 176 388 924 211 2;
  • 92) 0.205 709 176 388 924 211 2 × 2 = 0 + 0.411 418 352 777 848 422 4;
  • 93) 0.411 418 352 777 848 422 4 × 2 = 0 + 0.822 836 705 555 696 844 8;
  • 94) 0.822 836 705 555 696 844 8 × 2 = 1 + 0.645 673 411 111 393 689 6;
  • 95) 0.645 673 411 111 393 689 6 × 2 = 1 + 0.291 346 822 222 787 379 2;

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:


0.000 000 000 000 176 546 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1000 1011 0111 1011 0011 0110 1110 1100 0000 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 546 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1000 1011 0111 1011 0011 0110 1110 1100 0000 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:


0.000 000 000 000 176 546 9(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1000 1011 0111 1011 0011 0110 1110 1100 0000 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 1000 1011 0111 1011 0011 0110 1110 1100 0000 011(2) × 20 =

1.1000 1101 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011(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 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011


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 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011 =

1000 1101 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011


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 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011


Decimal number -0.000 000 000 000 176 546 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1000 1100 0101 1011 1101 1001 1011 0111 0110 0000 0011

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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