-0.000 000 000 000 176 557 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 557 68(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 68(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 68| = 0.000 000 000 000 176 557 68


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 68.

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 68 × 2 = 0 + 0.000 000 000 000 353 115 36;
  • 2) 0.000 000 000 000 353 115 36 × 2 = 0 + 0.000 000 000 000 706 230 72;
  • 3) 0.000 000 000 000 706 230 72 × 2 = 0 + 0.000 000 000 001 412 461 44;
  • 4) 0.000 000 000 001 412 461 44 × 2 = 0 + 0.000 000 000 002 824 922 88;
  • 5) 0.000 000 000 002 824 922 88 × 2 = 0 + 0.000 000 000 005 649 845 76;
  • 6) 0.000 000 000 005 649 845 76 × 2 = 0 + 0.000 000 000 011 299 691 52;
  • 7) 0.000 000 000 011 299 691 52 × 2 = 0 + 0.000 000 000 022 599 383 04;
  • 8) 0.000 000 000 022 599 383 04 × 2 = 0 + 0.000 000 000 045 198 766 08;
  • 9) 0.000 000 000 045 198 766 08 × 2 = 0 + 0.000 000 000 090 397 532 16;
  • 10) 0.000 000 000 090 397 532 16 × 2 = 0 + 0.000 000 000 180 795 064 32;
  • 11) 0.000 000 000 180 795 064 32 × 2 = 0 + 0.000 000 000 361 590 128 64;
  • 12) 0.000 000 000 361 590 128 64 × 2 = 0 + 0.000 000 000 723 180 257 28;
  • 13) 0.000 000 000 723 180 257 28 × 2 = 0 + 0.000 000 001 446 360 514 56;
  • 14) 0.000 000 001 446 360 514 56 × 2 = 0 + 0.000 000 002 892 721 029 12;
  • 15) 0.000 000 002 892 721 029 12 × 2 = 0 + 0.000 000 005 785 442 058 24;
  • 16) 0.000 000 005 785 442 058 24 × 2 = 0 + 0.000 000 011 570 884 116 48;
  • 17) 0.000 000 011 570 884 116 48 × 2 = 0 + 0.000 000 023 141 768 232 96;
  • 18) 0.000 000 023 141 768 232 96 × 2 = 0 + 0.000 000 046 283 536 465 92;
  • 19) 0.000 000 046 283 536 465 92 × 2 = 0 + 0.000 000 092 567 072 931 84;
  • 20) 0.000 000 092 567 072 931 84 × 2 = 0 + 0.000 000 185 134 145 863 68;
  • 21) 0.000 000 185 134 145 863 68 × 2 = 0 + 0.000 000 370 268 291 727 36;
  • 22) 0.000 000 370 268 291 727 36 × 2 = 0 + 0.000 000 740 536 583 454 72;
  • 23) 0.000 000 740 536 583 454 72 × 2 = 0 + 0.000 001 481 073 166 909 44;
  • 24) 0.000 001 481 073 166 909 44 × 2 = 0 + 0.000 002 962 146 333 818 88;
  • 25) 0.000 002 962 146 333 818 88 × 2 = 0 + 0.000 005 924 292 667 637 76;
  • 26) 0.000 005 924 292 667 637 76 × 2 = 0 + 0.000 011 848 585 335 275 52;
  • 27) 0.000 011 848 585 335 275 52 × 2 = 0 + 0.000 023 697 170 670 551 04;
  • 28) 0.000 023 697 170 670 551 04 × 2 = 0 + 0.000 047 394 341 341 102 08;
  • 29) 0.000 047 394 341 341 102 08 × 2 = 0 + 0.000 094 788 682 682 204 16;
  • 30) 0.000 094 788 682 682 204 16 × 2 = 0 + 0.000 189 577 365 364 408 32;
  • 31) 0.000 189 577 365 364 408 32 × 2 = 0 + 0.000 379 154 730 728 816 64;
  • 32) 0.000 379 154 730 728 816 64 × 2 = 0 + 0.000 758 309 461 457 633 28;
  • 33) 0.000 758 309 461 457 633 28 × 2 = 0 + 0.001 516 618 922 915 266 56;
  • 34) 0.001 516 618 922 915 266 56 × 2 = 0 + 0.003 033 237 845 830 533 12;
  • 35) 0.003 033 237 845 830 533 12 × 2 = 0 + 0.006 066 475 691 661 066 24;
  • 36) 0.006 066 475 691 661 066 24 × 2 = 0 + 0.012 132 951 383 322 132 48;
  • 37) 0.012 132 951 383 322 132 48 × 2 = 0 + 0.024 265 902 766 644 264 96;
  • 38) 0.024 265 902 766 644 264 96 × 2 = 0 + 0.048 531 805 533 288 529 92;
  • 39) 0.048 531 805 533 288 529 92 × 2 = 0 + 0.097 063 611 066 577 059 84;
  • 40) 0.097 063 611 066 577 059 84 × 2 = 0 + 0.194 127 222 133 154 119 68;
  • 41) 0.194 127 222 133 154 119 68 × 2 = 0 + 0.388 254 444 266 308 239 36;
  • 42) 0.388 254 444 266 308 239 36 × 2 = 0 + 0.776 508 888 532 616 478 72;
  • 43) 0.776 508 888 532 616 478 72 × 2 = 1 + 0.553 017 777 065 232 957 44;
  • 44) 0.553 017 777 065 232 957 44 × 2 = 1 + 0.106 035 554 130 465 914 88;
  • 45) 0.106 035 554 130 465 914 88 × 2 = 0 + 0.212 071 108 260 931 829 76;
  • 46) 0.212 071 108 260 931 829 76 × 2 = 0 + 0.424 142 216 521 863 659 52;
  • 47) 0.424 142 216 521 863 659 52 × 2 = 0 + 0.848 284 433 043 727 319 04;
  • 48) 0.848 284 433 043 727 319 04 × 2 = 1 + 0.696 568 866 087 454 638 08;
  • 49) 0.696 568 866 087 454 638 08 × 2 = 1 + 0.393 137 732 174 909 276 16;
  • 50) 0.393 137 732 174 909 276 16 × 2 = 0 + 0.786 275 464 349 818 552 32;
  • 51) 0.786 275 464 349 818 552 32 × 2 = 1 + 0.572 550 928 699 637 104 64;
  • 52) 0.572 550 928 699 637 104 64 × 2 = 1 + 0.145 101 857 399 274 209 28;
  • 53) 0.145 101 857 399 274 209 28 × 2 = 0 + 0.290 203 714 798 548 418 56;
  • 54) 0.290 203 714 798 548 418 56 × 2 = 0 + 0.580 407 429 597 096 837 12;
  • 55) 0.580 407 429 597 096 837 12 × 2 = 1 + 0.160 814 859 194 193 674 24;
  • 56) 0.160 814 859 194 193 674 24 × 2 = 0 + 0.321 629 718 388 387 348 48;
  • 57) 0.321 629 718 388 387 348 48 × 2 = 0 + 0.643 259 436 776 774 696 96;
  • 58) 0.643 259 436 776 774 696 96 × 2 = 1 + 0.286 518 873 553 549 393 92;
  • 59) 0.286 518 873 553 549 393 92 × 2 = 0 + 0.573 037 747 107 098 787 84;
  • 60) 0.573 037 747 107 098 787 84 × 2 = 1 + 0.146 075 494 214 197 575 68;
  • 61) 0.146 075 494 214 197 575 68 × 2 = 0 + 0.292 150 988 428 395 151 36;
  • 62) 0.292 150 988 428 395 151 36 × 2 = 0 + 0.584 301 976 856 790 302 72;
  • 63) 0.584 301 976 856 790 302 72 × 2 = 1 + 0.168 603 953 713 580 605 44;
  • 64) 0.168 603 953 713 580 605 44 × 2 = 0 + 0.337 207 907 427 161 210 88;
  • 65) 0.337 207 907 427 161 210 88 × 2 = 0 + 0.674 415 814 854 322 421 76;
  • 66) 0.674 415 814 854 322 421 76 × 2 = 1 + 0.348 831 629 708 644 843 52;
  • 67) 0.348 831 629 708 644 843 52 × 2 = 0 + 0.697 663 259 417 289 687 04;
  • 68) 0.697 663 259 417 289 687 04 × 2 = 1 + 0.395 326 518 834 579 374 08;
  • 69) 0.395 326 518 834 579 374 08 × 2 = 0 + 0.790 653 037 669 158 748 16;
  • 70) 0.790 653 037 669 158 748 16 × 2 = 1 + 0.581 306 075 338 317 496 32;
  • 71) 0.581 306 075 338 317 496 32 × 2 = 1 + 0.162 612 150 676 634 992 64;
  • 72) 0.162 612 150 676 634 992 64 × 2 = 0 + 0.325 224 301 353 269 985 28;
  • 73) 0.325 224 301 353 269 985 28 × 2 = 0 + 0.650 448 602 706 539 970 56;
  • 74) 0.650 448 602 706 539 970 56 × 2 = 1 + 0.300 897 205 413 079 941 12;
  • 75) 0.300 897 205 413 079 941 12 × 2 = 0 + 0.601 794 410 826 159 882 24;
  • 76) 0.601 794 410 826 159 882 24 × 2 = 1 + 0.203 588 821 652 319 764 48;
  • 77) 0.203 588 821 652 319 764 48 × 2 = 0 + 0.407 177 643 304 639 528 96;
  • 78) 0.407 177 643 304 639 528 96 × 2 = 0 + 0.814 355 286 609 279 057 92;
  • 79) 0.814 355 286 609 279 057 92 × 2 = 1 + 0.628 710 573 218 558 115 84;
  • 80) 0.628 710 573 218 558 115 84 × 2 = 1 + 0.257 421 146 437 116 231 68;
  • 81) 0.257 421 146 437 116 231 68 × 2 = 0 + 0.514 842 292 874 232 463 36;
  • 82) 0.514 842 292 874 232 463 36 × 2 = 1 + 0.029 684 585 748 464 926 72;
  • 83) 0.029 684 585 748 464 926 72 × 2 = 0 + 0.059 369 171 496 929 853 44;
  • 84) 0.059 369 171 496 929 853 44 × 2 = 0 + 0.118 738 342 993 859 706 88;
  • 85) 0.118 738 342 993 859 706 88 × 2 = 0 + 0.237 476 685 987 719 413 76;
  • 86) 0.237 476 685 987 719 413 76 × 2 = 0 + 0.474 953 371 975 438 827 52;
  • 87) 0.474 953 371 975 438 827 52 × 2 = 0 + 0.949 906 743 950 877 655 04;
  • 88) 0.949 906 743 950 877 655 04 × 2 = 1 + 0.899 813 487 901 755 310 08;
  • 89) 0.899 813 487 901 755 310 08 × 2 = 1 + 0.799 626 975 803 510 620 16;
  • 90) 0.799 626 975 803 510 620 16 × 2 = 1 + 0.599 253 951 607 021 240 32;
  • 91) 0.599 253 951 607 021 240 32 × 2 = 1 + 0.198 507 903 214 042 480 64;
  • 92) 0.198 507 903 214 042 480 64 × 2 = 0 + 0.397 015 806 428 084 961 28;
  • 93) 0.397 015 806 428 084 961 28 × 2 = 0 + 0.794 031 612 856 169 922 56;
  • 94) 0.794 031 612 856 169 922 56 × 2 = 1 + 0.588 063 225 712 339 845 12;
  • 95) 0.588 063 225 712 339 845 12 × 2 = 1 + 0.176 126 451 424 679 690 24;

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 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0101 0110 0101 0011 0100 0001 1110 011(2)

6. Positive number before normalization:

0.000 000 000 000 176 557 68(10) =

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


0.000 000 000 000 176 557 68(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0101 0110 0101 0011 0100 0001 1110 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 0010 0101 0110 0101 0011 0100 0001 1110 011(2) × 20 =

1.1000 1101 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 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 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 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 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 0011 =

1000 1101 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 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 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 0011


Decimal number -0.000 000 000 000 176 557 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 1001 0010 1011 0010 1001 1010 0000 1111 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