-0.000 000 000 000 176 558 03 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 558 03(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
-0.000 000 000 000 176 558 03(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:

|-0.000 000 000 000 176 558 03| = 0.000 000 000 000 176 558 03


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: 0.000 000 000 000 176 558 03.

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) 0.000 000 000 000 176 558 03 × 2 = 0 + 0.000 000 000 000 353 116 06;
  • 2) 0.000 000 000 000 353 116 06 × 2 = 0 + 0.000 000 000 000 706 232 12;
  • 3) 0.000 000 000 000 706 232 12 × 2 = 0 + 0.000 000 000 001 412 464 24;
  • 4) 0.000 000 000 001 412 464 24 × 2 = 0 + 0.000 000 000 002 824 928 48;
  • 5) 0.000 000 000 002 824 928 48 × 2 = 0 + 0.000 000 000 005 649 856 96;
  • 6) 0.000 000 000 005 649 856 96 × 2 = 0 + 0.000 000 000 011 299 713 92;
  • 7) 0.000 000 000 011 299 713 92 × 2 = 0 + 0.000 000 000 022 599 427 84;
  • 8) 0.000 000 000 022 599 427 84 × 2 = 0 + 0.000 000 000 045 198 855 68;
  • 9) 0.000 000 000 045 198 855 68 × 2 = 0 + 0.000 000 000 090 397 711 36;
  • 10) 0.000 000 000 090 397 711 36 × 2 = 0 + 0.000 000 000 180 795 422 72;
  • 11) 0.000 000 000 180 795 422 72 × 2 = 0 + 0.000 000 000 361 590 845 44;
  • 12) 0.000 000 000 361 590 845 44 × 2 = 0 + 0.000 000 000 723 181 690 88;
  • 13) 0.000 000 000 723 181 690 88 × 2 = 0 + 0.000 000 001 446 363 381 76;
  • 14) 0.000 000 001 446 363 381 76 × 2 = 0 + 0.000 000 002 892 726 763 52;
  • 15) 0.000 000 002 892 726 763 52 × 2 = 0 + 0.000 000 005 785 453 527 04;
  • 16) 0.000 000 005 785 453 527 04 × 2 = 0 + 0.000 000 011 570 907 054 08;
  • 17) 0.000 000 011 570 907 054 08 × 2 = 0 + 0.000 000 023 141 814 108 16;
  • 18) 0.000 000 023 141 814 108 16 × 2 = 0 + 0.000 000 046 283 628 216 32;
  • 19) 0.000 000 046 283 628 216 32 × 2 = 0 + 0.000 000 092 567 256 432 64;
  • 20) 0.000 000 092 567 256 432 64 × 2 = 0 + 0.000 000 185 134 512 865 28;
  • 21) 0.000 000 185 134 512 865 28 × 2 = 0 + 0.000 000 370 269 025 730 56;
  • 22) 0.000 000 370 269 025 730 56 × 2 = 0 + 0.000 000 740 538 051 461 12;
  • 23) 0.000 000 740 538 051 461 12 × 2 = 0 + 0.000 001 481 076 102 922 24;
  • 24) 0.000 001 481 076 102 922 24 × 2 = 0 + 0.000 002 962 152 205 844 48;
  • 25) 0.000 002 962 152 205 844 48 × 2 = 0 + 0.000 005 924 304 411 688 96;
  • 26) 0.000 005 924 304 411 688 96 × 2 = 0 + 0.000 011 848 608 823 377 92;
  • 27) 0.000 011 848 608 823 377 92 × 2 = 0 + 0.000 023 697 217 646 755 84;
  • 28) 0.000 023 697 217 646 755 84 × 2 = 0 + 0.000 047 394 435 293 511 68;
  • 29) 0.000 047 394 435 293 511 68 × 2 = 0 + 0.000 094 788 870 587 023 36;
  • 30) 0.000 094 788 870 587 023 36 × 2 = 0 + 0.000 189 577 741 174 046 72;
  • 31) 0.000 189 577 741 174 046 72 × 2 = 0 + 0.000 379 155 482 348 093 44;
  • 32) 0.000 379 155 482 348 093 44 × 2 = 0 + 0.000 758 310 964 696 186 88;
  • 33) 0.000 758 310 964 696 186 88 × 2 = 0 + 0.001 516 621 929 392 373 76;
  • 34) 0.001 516 621 929 392 373 76 × 2 = 0 + 0.003 033 243 858 784 747 52;
  • 35) 0.003 033 243 858 784 747 52 × 2 = 0 + 0.006 066 487 717 569 495 04;
  • 36) 0.006 066 487 717 569 495 04 × 2 = 0 + 0.012 132 975 435 138 990 08;
  • 37) 0.012 132 975 435 138 990 08 × 2 = 0 + 0.024 265 950 870 277 980 16;
  • 38) 0.024 265 950 870 277 980 16 × 2 = 0 + 0.048 531 901 740 555 960 32;
  • 39) 0.048 531 901 740 555 960 32 × 2 = 0 + 0.097 063 803 481 111 920 64;
  • 40) 0.097 063 803 481 111 920 64 × 2 = 0 + 0.194 127 606 962 223 841 28;
  • 41) 0.194 127 606 962 223 841 28 × 2 = 0 + 0.388 255 213 924 447 682 56;
  • 42) 0.388 255 213 924 447 682 56 × 2 = 0 + 0.776 510 427 848 895 365 12;
  • 43) 0.776 510 427 848 895 365 12 × 2 = 1 + 0.553 020 855 697 790 730 24;
  • 44) 0.553 020 855 697 790 730 24 × 2 = 1 + 0.106 041 711 395 581 460 48;
  • 45) 0.106 041 711 395 581 460 48 × 2 = 0 + 0.212 083 422 791 162 920 96;
  • 46) 0.212 083 422 791 162 920 96 × 2 = 0 + 0.424 166 845 582 325 841 92;
  • 47) 0.424 166 845 582 325 841 92 × 2 = 0 + 0.848 333 691 164 651 683 84;
  • 48) 0.848 333 691 164 651 683 84 × 2 = 1 + 0.696 667 382 329 303 367 68;
  • 49) 0.696 667 382 329 303 367 68 × 2 = 1 + 0.393 334 764 658 606 735 36;
  • 50) 0.393 334 764 658 606 735 36 × 2 = 0 + 0.786 669 529 317 213 470 72;
  • 51) 0.786 669 529 317 213 470 72 × 2 = 1 + 0.573 339 058 634 426 941 44;
  • 52) 0.573 339 058 634 426 941 44 × 2 = 1 + 0.146 678 117 268 853 882 88;
  • 53) 0.146 678 117 268 853 882 88 × 2 = 0 + 0.293 356 234 537 707 765 76;
  • 54) 0.293 356 234 537 707 765 76 × 2 = 0 + 0.586 712 469 075 415 531 52;
  • 55) 0.586 712 469 075 415 531 52 × 2 = 1 + 0.173 424 938 150 831 063 04;
  • 56) 0.173 424 938 150 831 063 04 × 2 = 0 + 0.346 849 876 301 662 126 08;
  • 57) 0.346 849 876 301 662 126 08 × 2 = 0 + 0.693 699 752 603 324 252 16;
  • 58) 0.693 699 752 603 324 252 16 × 2 = 1 + 0.387 399 505 206 648 504 32;
  • 59) 0.387 399 505 206 648 504 32 × 2 = 0 + 0.774 799 010 413 297 008 64;
  • 60) 0.774 799 010 413 297 008 64 × 2 = 1 + 0.549 598 020 826 594 017 28;
  • 61) 0.549 598 020 826 594 017 28 × 2 = 1 + 0.099 196 041 653 188 034 56;
  • 62) 0.099 196 041 653 188 034 56 × 2 = 0 + 0.198 392 083 306 376 069 12;
  • 63) 0.198 392 083 306 376 069 12 × 2 = 0 + 0.396 784 166 612 752 138 24;
  • 64) 0.396 784 166 612 752 138 24 × 2 = 0 + 0.793 568 333 225 504 276 48;
  • 65) 0.793 568 333 225 504 276 48 × 2 = 1 + 0.587 136 666 451 008 552 96;
  • 66) 0.587 136 666 451 008 552 96 × 2 = 1 + 0.174 273 332 902 017 105 92;
  • 67) 0.174 273 332 902 017 105 92 × 2 = 0 + 0.348 546 665 804 034 211 84;
  • 68) 0.348 546 665 804 034 211 84 × 2 = 0 + 0.697 093 331 608 068 423 68;
  • 69) 0.697 093 331 608 068 423 68 × 2 = 1 + 0.394 186 663 216 136 847 36;
  • 70) 0.394 186 663 216 136 847 36 × 2 = 0 + 0.788 373 326 432 273 694 72;
  • 71) 0.788 373 326 432 273 694 72 × 2 = 1 + 0.576 746 652 864 547 389 44;
  • 72) 0.576 746 652 864 547 389 44 × 2 = 1 + 0.153 493 305 729 094 778 88;
  • 73) 0.153 493 305 729 094 778 88 × 2 = 0 + 0.306 986 611 458 189 557 76;
  • 74) 0.306 986 611 458 189 557 76 × 2 = 0 + 0.613 973 222 916 379 115 52;
  • 75) 0.613 973 222 916 379 115 52 × 2 = 1 + 0.227 946 445 832 758 231 04;
  • 76) 0.227 946 445 832 758 231 04 × 2 = 0 + 0.455 892 891 665 516 462 08;
  • 77) 0.455 892 891 665 516 462 08 × 2 = 0 + 0.911 785 783 331 032 924 16;
  • 78) 0.911 785 783 331 032 924 16 × 2 = 1 + 0.823 571 566 662 065 848 32;
  • 79) 0.823 571 566 662 065 848 32 × 2 = 1 + 0.647 143 133 324 131 696 64;
  • 80) 0.647 143 133 324 131 696 64 × 2 = 1 + 0.294 286 266 648 263 393 28;
  • 81) 0.294 286 266 648 263 393 28 × 2 = 0 + 0.588 572 533 296 526 786 56;
  • 82) 0.588 572 533 296 526 786 56 × 2 = 1 + 0.177 145 066 593 053 573 12;
  • 83) 0.177 145 066 593 053 573 12 × 2 = 0 + 0.354 290 133 186 107 146 24;
  • 84) 0.354 290 133 186 107 146 24 × 2 = 0 + 0.708 580 266 372 214 292 48;
  • 85) 0.708 580 266 372 214 292 48 × 2 = 1 + 0.417 160 532 744 428 584 96;
  • 86) 0.417 160 532 744 428 584 96 × 2 = 0 + 0.834 321 065 488 857 169 92;
  • 87) 0.834 321 065 488 857 169 92 × 2 = 1 + 0.668 642 130 977 714 339 84;
  • 88) 0.668 642 130 977 714 339 84 × 2 = 1 + 0.337 284 261 955 428 679 68;
  • 89) 0.337 284 261 955 428 679 68 × 2 = 0 + 0.674 568 523 910 857 359 36;
  • 90) 0.674 568 523 910 857 359 36 × 2 = 1 + 0.349 137 047 821 714 718 72;
  • 91) 0.349 137 047 821 714 718 72 × 2 = 0 + 0.698 274 095 643 429 437 44;
  • 92) 0.698 274 095 643 429 437 44 × 2 = 1 + 0.396 548 191 286 858 874 88;
  • 93) 0.396 548 191 286 858 874 88 × 2 = 0 + 0.793 096 382 573 717 749 76;
  • 94) 0.793 096 382 573 717 749 76 × 2 = 1 + 0.586 192 765 147 435 499 52;
  • 95) 0.586 192 765 147 435 499 52 × 2 = 1 + 0.172 385 530 294 870 999 04;

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:


0.000 000 000 000 176 558 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1000 1100 1011 0010 0111 0100 1011 0101 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 558 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1000 1100 1011 0010 0111 0100 1011 0101 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:


0.000 000 000 000 176 558 03(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1000 1100 1011 0010 0111 0100 1011 0101 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1000 1100 1011 0010 0111 0100 1011 0101 011(2) × 20 =

1.1000 1101 1001 0010 1100 0110 0101 1001 0011 1010 0101 1010 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 1100 0110 0101 1001 0011 1010 0101 1010 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 1100 0110 0101 1001 0011 1010 0101 1010 1011 =

1000 1101 1001 0010 1100 0110 0101 1001 0011 1010 0101 1010 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 1100 0110 0101 1001 0011 1010 0101 1010 1011


Decimal number -0.000 000 000 000 176 558 03 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1100 0110 0101 1001 0011 1010 0101 1010 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