-0.000 000 000 000 176 557 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 2(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 2(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 2| = 0.000 000 000 000 176 557 2


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 2.

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 2 × 2 = 0 + 0.000 000 000 000 353 114 4;
  • 2) 0.000 000 000 000 353 114 4 × 2 = 0 + 0.000 000 000 000 706 228 8;
  • 3) 0.000 000 000 000 706 228 8 × 2 = 0 + 0.000 000 000 001 412 457 6;
  • 4) 0.000 000 000 001 412 457 6 × 2 = 0 + 0.000 000 000 002 824 915 2;
  • 5) 0.000 000 000 002 824 915 2 × 2 = 0 + 0.000 000 000 005 649 830 4;
  • 6) 0.000 000 000 005 649 830 4 × 2 = 0 + 0.000 000 000 011 299 660 8;
  • 7) 0.000 000 000 011 299 660 8 × 2 = 0 + 0.000 000 000 022 599 321 6;
  • 8) 0.000 000 000 022 599 321 6 × 2 = 0 + 0.000 000 000 045 198 643 2;
  • 9) 0.000 000 000 045 198 643 2 × 2 = 0 + 0.000 000 000 090 397 286 4;
  • 10) 0.000 000 000 090 397 286 4 × 2 = 0 + 0.000 000 000 180 794 572 8;
  • 11) 0.000 000 000 180 794 572 8 × 2 = 0 + 0.000 000 000 361 589 145 6;
  • 12) 0.000 000 000 361 589 145 6 × 2 = 0 + 0.000 000 000 723 178 291 2;
  • 13) 0.000 000 000 723 178 291 2 × 2 = 0 + 0.000 000 001 446 356 582 4;
  • 14) 0.000 000 001 446 356 582 4 × 2 = 0 + 0.000 000 002 892 713 164 8;
  • 15) 0.000 000 002 892 713 164 8 × 2 = 0 + 0.000 000 005 785 426 329 6;
  • 16) 0.000 000 005 785 426 329 6 × 2 = 0 + 0.000 000 011 570 852 659 2;
  • 17) 0.000 000 011 570 852 659 2 × 2 = 0 + 0.000 000 023 141 705 318 4;
  • 18) 0.000 000 023 141 705 318 4 × 2 = 0 + 0.000 000 046 283 410 636 8;
  • 19) 0.000 000 046 283 410 636 8 × 2 = 0 + 0.000 000 092 566 821 273 6;
  • 20) 0.000 000 092 566 821 273 6 × 2 = 0 + 0.000 000 185 133 642 547 2;
  • 21) 0.000 000 185 133 642 547 2 × 2 = 0 + 0.000 000 370 267 285 094 4;
  • 22) 0.000 000 370 267 285 094 4 × 2 = 0 + 0.000 000 740 534 570 188 8;
  • 23) 0.000 000 740 534 570 188 8 × 2 = 0 + 0.000 001 481 069 140 377 6;
  • 24) 0.000 001 481 069 140 377 6 × 2 = 0 + 0.000 002 962 138 280 755 2;
  • 25) 0.000 002 962 138 280 755 2 × 2 = 0 + 0.000 005 924 276 561 510 4;
  • 26) 0.000 005 924 276 561 510 4 × 2 = 0 + 0.000 011 848 553 123 020 8;
  • 27) 0.000 011 848 553 123 020 8 × 2 = 0 + 0.000 023 697 106 246 041 6;
  • 28) 0.000 023 697 106 246 041 6 × 2 = 0 + 0.000 047 394 212 492 083 2;
  • 29) 0.000 047 394 212 492 083 2 × 2 = 0 + 0.000 094 788 424 984 166 4;
  • 30) 0.000 094 788 424 984 166 4 × 2 = 0 + 0.000 189 576 849 968 332 8;
  • 31) 0.000 189 576 849 968 332 8 × 2 = 0 + 0.000 379 153 699 936 665 6;
  • 32) 0.000 379 153 699 936 665 6 × 2 = 0 + 0.000 758 307 399 873 331 2;
  • 33) 0.000 758 307 399 873 331 2 × 2 = 0 + 0.001 516 614 799 746 662 4;
  • 34) 0.001 516 614 799 746 662 4 × 2 = 0 + 0.003 033 229 599 493 324 8;
  • 35) 0.003 033 229 599 493 324 8 × 2 = 0 + 0.006 066 459 198 986 649 6;
  • 36) 0.006 066 459 198 986 649 6 × 2 = 0 + 0.012 132 918 397 973 299 2;
  • 37) 0.012 132 918 397 973 299 2 × 2 = 0 + 0.024 265 836 795 946 598 4;
  • 38) 0.024 265 836 795 946 598 4 × 2 = 0 + 0.048 531 673 591 893 196 8;
  • 39) 0.048 531 673 591 893 196 8 × 2 = 0 + 0.097 063 347 183 786 393 6;
  • 40) 0.097 063 347 183 786 393 6 × 2 = 0 + 0.194 126 694 367 572 787 2;
  • 41) 0.194 126 694 367 572 787 2 × 2 = 0 + 0.388 253 388 735 145 574 4;
  • 42) 0.388 253 388 735 145 574 4 × 2 = 0 + 0.776 506 777 470 291 148 8;
  • 43) 0.776 506 777 470 291 148 8 × 2 = 1 + 0.553 013 554 940 582 297 6;
  • 44) 0.553 013 554 940 582 297 6 × 2 = 1 + 0.106 027 109 881 164 595 2;
  • 45) 0.106 027 109 881 164 595 2 × 2 = 0 + 0.212 054 219 762 329 190 4;
  • 46) 0.212 054 219 762 329 190 4 × 2 = 0 + 0.424 108 439 524 658 380 8;
  • 47) 0.424 108 439 524 658 380 8 × 2 = 0 + 0.848 216 879 049 316 761 6;
  • 48) 0.848 216 879 049 316 761 6 × 2 = 1 + 0.696 433 758 098 633 523 2;
  • 49) 0.696 433 758 098 633 523 2 × 2 = 1 + 0.392 867 516 197 267 046 4;
  • 50) 0.392 867 516 197 267 046 4 × 2 = 0 + 0.785 735 032 394 534 092 8;
  • 51) 0.785 735 032 394 534 092 8 × 2 = 1 + 0.571 470 064 789 068 185 6;
  • 52) 0.571 470 064 789 068 185 6 × 2 = 1 + 0.142 940 129 578 136 371 2;
  • 53) 0.142 940 129 578 136 371 2 × 2 = 0 + 0.285 880 259 156 272 742 4;
  • 54) 0.285 880 259 156 272 742 4 × 2 = 0 + 0.571 760 518 312 545 484 8;
  • 55) 0.571 760 518 312 545 484 8 × 2 = 1 + 0.143 521 036 625 090 969 6;
  • 56) 0.143 521 036 625 090 969 6 × 2 = 0 + 0.287 042 073 250 181 939 2;
  • 57) 0.287 042 073 250 181 939 2 × 2 = 0 + 0.574 084 146 500 363 878 4;
  • 58) 0.574 084 146 500 363 878 4 × 2 = 1 + 0.148 168 293 000 727 756 8;
  • 59) 0.148 168 293 000 727 756 8 × 2 = 0 + 0.296 336 586 001 455 513 6;
  • 60) 0.296 336 586 001 455 513 6 × 2 = 0 + 0.592 673 172 002 911 027 2;
  • 61) 0.592 673 172 002 911 027 2 × 2 = 1 + 0.185 346 344 005 822 054 4;
  • 62) 0.185 346 344 005 822 054 4 × 2 = 0 + 0.370 692 688 011 644 108 8;
  • 63) 0.370 692 688 011 644 108 8 × 2 = 0 + 0.741 385 376 023 288 217 6;
  • 64) 0.741 385 376 023 288 217 6 × 2 = 1 + 0.482 770 752 046 576 435 2;
  • 65) 0.482 770 752 046 576 435 2 × 2 = 0 + 0.965 541 504 093 152 870 4;
  • 66) 0.965 541 504 093 152 870 4 × 2 = 1 + 0.931 083 008 186 305 740 8;
  • 67) 0.931 083 008 186 305 740 8 × 2 = 1 + 0.862 166 016 372 611 481 6;
  • 68) 0.862 166 016 372 611 481 6 × 2 = 1 + 0.724 332 032 745 222 963 2;
  • 69) 0.724 332 032 745 222 963 2 × 2 = 1 + 0.448 664 065 490 445 926 4;
  • 70) 0.448 664 065 490 445 926 4 × 2 = 0 + 0.897 328 130 980 891 852 8;
  • 71) 0.897 328 130 980 891 852 8 × 2 = 1 + 0.794 656 261 961 783 705 6;
  • 72) 0.794 656 261 961 783 705 6 × 2 = 1 + 0.589 312 523 923 567 411 2;
  • 73) 0.589 312 523 923 567 411 2 × 2 = 1 + 0.178 625 047 847 134 822 4;
  • 74) 0.178 625 047 847 134 822 4 × 2 = 0 + 0.357 250 095 694 269 644 8;
  • 75) 0.357 250 095 694 269 644 8 × 2 = 0 + 0.714 500 191 388 539 289 6;
  • 76) 0.714 500 191 388 539 289 6 × 2 = 1 + 0.429 000 382 777 078 579 2;
  • 77) 0.429 000 382 777 078 579 2 × 2 = 0 + 0.858 000 765 554 157 158 4;
  • 78) 0.858 000 765 554 157 158 4 × 2 = 1 + 0.716 001 531 108 314 316 8;
  • 79) 0.716 001 531 108 314 316 8 × 2 = 1 + 0.432 003 062 216 628 633 6;
  • 80) 0.432 003 062 216 628 633 6 × 2 = 0 + 0.864 006 124 433 257 267 2;
  • 81) 0.864 006 124 433 257 267 2 × 2 = 1 + 0.728 012 248 866 514 534 4;
  • 82) 0.728 012 248 866 514 534 4 × 2 = 1 + 0.456 024 497 733 029 068 8;
  • 83) 0.456 024 497 733 029 068 8 × 2 = 0 + 0.912 048 995 466 058 137 6;
  • 84) 0.912 048 995 466 058 137 6 × 2 = 1 + 0.824 097 990 932 116 275 2;
  • 85) 0.824 097 990 932 116 275 2 × 2 = 1 + 0.648 195 981 864 232 550 4;
  • 86) 0.648 195 981 864 232 550 4 × 2 = 1 + 0.296 391 963 728 465 100 8;
  • 87) 0.296 391 963 728 465 100 8 × 2 = 0 + 0.592 783 927 456 930 201 6;
  • 88) 0.592 783 927 456 930 201 6 × 2 = 1 + 0.185 567 854 913 860 403 2;
  • 89) 0.185 567 854 913 860 403 2 × 2 = 0 + 0.371 135 709 827 720 806 4;
  • 90) 0.371 135 709 827 720 806 4 × 2 = 0 + 0.742 271 419 655 441 612 8;
  • 91) 0.742 271 419 655 441 612 8 × 2 = 1 + 0.484 542 839 310 883 225 6;
  • 92) 0.484 542 839 310 883 225 6 × 2 = 0 + 0.969 085 678 621 766 451 2;
  • 93) 0.969 085 678 621 766 451 2 × 2 = 1 + 0.938 171 357 243 532 902 4;
  • 94) 0.938 171 357 243 532 902 4 × 2 = 1 + 0.876 342 714 487 065 804 8;
  • 95) 0.876 342 714 487 065 804 8 × 2 = 1 + 0.752 685 428 974 131 609 6;

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 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0111 1011 1001 0110 1101 1101 0010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0111 1011 1001 0110 1101 1101 0010 111(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 2(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0111 1011 1001 0110 1101 1101 0010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 1001 0111 1011 1001 0110 1101 1101 0010 111(2) × 20 =

1.1000 1101 1001 0010 0100 1011 1101 1100 1011 0110 1110 1001 0111(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 1101 1100 1011 0110 1110 1001 0111


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 1101 1100 1011 0110 1110 1001 0111 =

1000 1101 1001 0010 0100 1011 1101 1100 1011 0110 1110 1001 0111


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 1101 1100 1011 0110 1110 1001 0111


Decimal number -0.000 000 000 000 176 557 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0100 1011 1101 1100 1011 0110 1110 1001 0111

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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