-0.000 000 000 000 176 555 73 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 555 73(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
-0.000 000 000 000 176 555 73(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:

|-0.000 000 000 000 176 555 73| = 0.000 000 000 000 176 555 73


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: 0.000 000 000 000 176 555 73.

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) 0.000 000 000 000 176 555 73 × 2 = 0 + 0.000 000 000 000 353 111 46;
  • 2) 0.000 000 000 000 353 111 46 × 2 = 0 + 0.000 000 000 000 706 222 92;
  • 3) 0.000 000 000 000 706 222 92 × 2 = 0 + 0.000 000 000 001 412 445 84;
  • 4) 0.000 000 000 001 412 445 84 × 2 = 0 + 0.000 000 000 002 824 891 68;
  • 5) 0.000 000 000 002 824 891 68 × 2 = 0 + 0.000 000 000 005 649 783 36;
  • 6) 0.000 000 000 005 649 783 36 × 2 = 0 + 0.000 000 000 011 299 566 72;
  • 7) 0.000 000 000 011 299 566 72 × 2 = 0 + 0.000 000 000 022 599 133 44;
  • 8) 0.000 000 000 022 599 133 44 × 2 = 0 + 0.000 000 000 045 198 266 88;
  • 9) 0.000 000 000 045 198 266 88 × 2 = 0 + 0.000 000 000 090 396 533 76;
  • 10) 0.000 000 000 090 396 533 76 × 2 = 0 + 0.000 000 000 180 793 067 52;
  • 11) 0.000 000 000 180 793 067 52 × 2 = 0 + 0.000 000 000 361 586 135 04;
  • 12) 0.000 000 000 361 586 135 04 × 2 = 0 + 0.000 000 000 723 172 270 08;
  • 13) 0.000 000 000 723 172 270 08 × 2 = 0 + 0.000 000 001 446 344 540 16;
  • 14) 0.000 000 001 446 344 540 16 × 2 = 0 + 0.000 000 002 892 689 080 32;
  • 15) 0.000 000 002 892 689 080 32 × 2 = 0 + 0.000 000 005 785 378 160 64;
  • 16) 0.000 000 005 785 378 160 64 × 2 = 0 + 0.000 000 011 570 756 321 28;
  • 17) 0.000 000 011 570 756 321 28 × 2 = 0 + 0.000 000 023 141 512 642 56;
  • 18) 0.000 000 023 141 512 642 56 × 2 = 0 + 0.000 000 046 283 025 285 12;
  • 19) 0.000 000 046 283 025 285 12 × 2 = 0 + 0.000 000 092 566 050 570 24;
  • 20) 0.000 000 092 566 050 570 24 × 2 = 0 + 0.000 000 185 132 101 140 48;
  • 21) 0.000 000 185 132 101 140 48 × 2 = 0 + 0.000 000 370 264 202 280 96;
  • 22) 0.000 000 370 264 202 280 96 × 2 = 0 + 0.000 000 740 528 404 561 92;
  • 23) 0.000 000 740 528 404 561 92 × 2 = 0 + 0.000 001 481 056 809 123 84;
  • 24) 0.000 001 481 056 809 123 84 × 2 = 0 + 0.000 002 962 113 618 247 68;
  • 25) 0.000 002 962 113 618 247 68 × 2 = 0 + 0.000 005 924 227 236 495 36;
  • 26) 0.000 005 924 227 236 495 36 × 2 = 0 + 0.000 011 848 454 472 990 72;
  • 27) 0.000 011 848 454 472 990 72 × 2 = 0 + 0.000 023 696 908 945 981 44;
  • 28) 0.000 023 696 908 945 981 44 × 2 = 0 + 0.000 047 393 817 891 962 88;
  • 29) 0.000 047 393 817 891 962 88 × 2 = 0 + 0.000 094 787 635 783 925 76;
  • 30) 0.000 094 787 635 783 925 76 × 2 = 0 + 0.000 189 575 271 567 851 52;
  • 31) 0.000 189 575 271 567 851 52 × 2 = 0 + 0.000 379 150 543 135 703 04;
  • 32) 0.000 379 150 543 135 703 04 × 2 = 0 + 0.000 758 301 086 271 406 08;
  • 33) 0.000 758 301 086 271 406 08 × 2 = 0 + 0.001 516 602 172 542 812 16;
  • 34) 0.001 516 602 172 542 812 16 × 2 = 0 + 0.003 033 204 345 085 624 32;
  • 35) 0.003 033 204 345 085 624 32 × 2 = 0 + 0.006 066 408 690 171 248 64;
  • 36) 0.006 066 408 690 171 248 64 × 2 = 0 + 0.012 132 817 380 342 497 28;
  • 37) 0.012 132 817 380 342 497 28 × 2 = 0 + 0.024 265 634 760 684 994 56;
  • 38) 0.024 265 634 760 684 994 56 × 2 = 0 + 0.048 531 269 521 369 989 12;
  • 39) 0.048 531 269 521 369 989 12 × 2 = 0 + 0.097 062 539 042 739 978 24;
  • 40) 0.097 062 539 042 739 978 24 × 2 = 0 + 0.194 125 078 085 479 956 48;
  • 41) 0.194 125 078 085 479 956 48 × 2 = 0 + 0.388 250 156 170 959 912 96;
  • 42) 0.388 250 156 170 959 912 96 × 2 = 0 + 0.776 500 312 341 919 825 92;
  • 43) 0.776 500 312 341 919 825 92 × 2 = 1 + 0.553 000 624 683 839 651 84;
  • 44) 0.553 000 624 683 839 651 84 × 2 = 1 + 0.106 001 249 367 679 303 68;
  • 45) 0.106 001 249 367 679 303 68 × 2 = 0 + 0.212 002 498 735 358 607 36;
  • 46) 0.212 002 498 735 358 607 36 × 2 = 0 + 0.424 004 997 470 717 214 72;
  • 47) 0.424 004 997 470 717 214 72 × 2 = 0 + 0.848 009 994 941 434 429 44;
  • 48) 0.848 009 994 941 434 429 44 × 2 = 1 + 0.696 019 989 882 868 858 88;
  • 49) 0.696 019 989 882 868 858 88 × 2 = 1 + 0.392 039 979 765 737 717 76;
  • 50) 0.392 039 979 765 737 717 76 × 2 = 0 + 0.784 079 959 531 475 435 52;
  • 51) 0.784 079 959 531 475 435 52 × 2 = 1 + 0.568 159 919 062 950 871 04;
  • 52) 0.568 159 919 062 950 871 04 × 2 = 1 + 0.136 319 838 125 901 742 08;
  • 53) 0.136 319 838 125 901 742 08 × 2 = 0 + 0.272 639 676 251 803 484 16;
  • 54) 0.272 639 676 251 803 484 16 × 2 = 0 + 0.545 279 352 503 606 968 32;
  • 55) 0.545 279 352 503 606 968 32 × 2 = 1 + 0.090 558 705 007 213 936 64;
  • 56) 0.090 558 705 007 213 936 64 × 2 = 0 + 0.181 117 410 014 427 873 28;
  • 57) 0.181 117 410 014 427 873 28 × 2 = 0 + 0.362 234 820 028 855 746 56;
  • 58) 0.362 234 820 028 855 746 56 × 2 = 0 + 0.724 469 640 057 711 493 12;
  • 59) 0.724 469 640 057 711 493 12 × 2 = 1 + 0.448 939 280 115 422 986 24;
  • 60) 0.448 939 280 115 422 986 24 × 2 = 0 + 0.897 878 560 230 845 972 48;
  • 61) 0.897 878 560 230 845 972 48 × 2 = 1 + 0.795 757 120 461 691 944 96;
  • 62) 0.795 757 120 461 691 944 96 × 2 = 1 + 0.591 514 240 923 383 889 92;
  • 63) 0.591 514 240 923 383 889 92 × 2 = 1 + 0.183 028 481 846 767 779 84;
  • 64) 0.183 028 481 846 767 779 84 × 2 = 0 + 0.366 056 963 693 535 559 68;
  • 65) 0.366 056 963 693 535 559 68 × 2 = 0 + 0.732 113 927 387 071 119 36;
  • 66) 0.732 113 927 387 071 119 36 × 2 = 1 + 0.464 227 854 774 142 238 72;
  • 67) 0.464 227 854 774 142 238 72 × 2 = 0 + 0.928 455 709 548 284 477 44;
  • 68) 0.928 455 709 548 284 477 44 × 2 = 1 + 0.856 911 419 096 568 954 88;
  • 69) 0.856 911 419 096 568 954 88 × 2 = 1 + 0.713 822 838 193 137 909 76;
  • 70) 0.713 822 838 193 137 909 76 × 2 = 1 + 0.427 645 676 386 275 819 52;
  • 71) 0.427 645 676 386 275 819 52 × 2 = 0 + 0.855 291 352 772 551 639 04;
  • 72) 0.855 291 352 772 551 639 04 × 2 = 1 + 0.710 582 705 545 103 278 08;
  • 73) 0.710 582 705 545 103 278 08 × 2 = 1 + 0.421 165 411 090 206 556 16;
  • 74) 0.421 165 411 090 206 556 16 × 2 = 0 + 0.842 330 822 180 413 112 32;
  • 75) 0.842 330 822 180 413 112 32 × 2 = 1 + 0.684 661 644 360 826 224 64;
  • 76) 0.684 661 644 360 826 224 64 × 2 = 1 + 0.369 323 288 721 652 449 28;
  • 77) 0.369 323 288 721 652 449 28 × 2 = 0 + 0.738 646 577 443 304 898 56;
  • 78) 0.738 646 577 443 304 898 56 × 2 = 1 + 0.477 293 154 886 609 797 12;
  • 79) 0.477 293 154 886 609 797 12 × 2 = 0 + 0.954 586 309 773 219 594 24;
  • 80) 0.954 586 309 773 219 594 24 × 2 = 1 + 0.909 172 619 546 439 188 48;
  • 81) 0.909 172 619 546 439 188 48 × 2 = 1 + 0.818 345 239 092 878 376 96;
  • 82) 0.818 345 239 092 878 376 96 × 2 = 1 + 0.636 690 478 185 756 753 92;
  • 83) 0.636 690 478 185 756 753 92 × 2 = 1 + 0.273 380 956 371 513 507 84;
  • 84) 0.273 380 956 371 513 507 84 × 2 = 0 + 0.546 761 912 743 027 015 68;
  • 85) 0.546 761 912 743 027 015 68 × 2 = 1 + 0.093 523 825 486 054 031 36;
  • 86) 0.093 523 825 486 054 031 36 × 2 = 0 + 0.187 047 650 972 108 062 72;
  • 87) 0.187 047 650 972 108 062 72 × 2 = 0 + 0.374 095 301 944 216 125 44;
  • 88) 0.374 095 301 944 216 125 44 × 2 = 0 + 0.748 190 603 888 432 250 88;
  • 89) 0.748 190 603 888 432 250 88 × 2 = 1 + 0.496 381 207 776 864 501 76;
  • 90) 0.496 381 207 776 864 501 76 × 2 = 0 + 0.992 762 415 553 729 003 52;
  • 91) 0.992 762 415 553 729 003 52 × 2 = 1 + 0.985 524 831 107 458 007 04;
  • 92) 0.985 524 831 107 458 007 04 × 2 = 1 + 0.971 049 662 214 916 014 08;
  • 93) 0.971 049 662 214 916 014 08 × 2 = 1 + 0.942 099 324 429 832 028 16;
  • 94) 0.942 099 324 429 832 028 16 × 2 = 1 + 0.884 198 648 859 664 056 32;
  • 95) 0.884 198 648 859 664 056 32 × 2 = 1 + 0.768 397 297 719 328 112 64;

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:


0.000 000 000 000 176 555 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1110 0101 1101 1011 0101 1110 1000 1011 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 555 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1110 0101 1101 1011 0101 1110 1000 1011 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:


0.000 000 000 000 176 555 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1110 0101 1101 1011 0101 1110 1000 1011 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1110 0101 1101 1011 0101 1110 1000 1011 111(2) × 20 =

1.1000 1101 1001 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111(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 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111


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 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111 =

1000 1101 1001 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111


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 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111


Decimal number -0.000 000 000 000 176 555 73 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 0111 0010 1110 1101 1010 1111 0100 0101 1111

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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