-0.000 000 000 000 176 556 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 64(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
-0.000 000 000 000 176 556 64(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:

|-0.000 000 000 000 176 556 64| = 0.000 000 000 000 176 556 64


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: 0.000 000 000 000 176 556 64.

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) 0.000 000 000 000 176 556 64 × 2 = 0 + 0.000 000 000 000 353 113 28;
  • 2) 0.000 000 000 000 353 113 28 × 2 = 0 + 0.000 000 000 000 706 226 56;
  • 3) 0.000 000 000 000 706 226 56 × 2 = 0 + 0.000 000 000 001 412 453 12;
  • 4) 0.000 000 000 001 412 453 12 × 2 = 0 + 0.000 000 000 002 824 906 24;
  • 5) 0.000 000 000 002 824 906 24 × 2 = 0 + 0.000 000 000 005 649 812 48;
  • 6) 0.000 000 000 005 649 812 48 × 2 = 0 + 0.000 000 000 011 299 624 96;
  • 7) 0.000 000 000 011 299 624 96 × 2 = 0 + 0.000 000 000 022 599 249 92;
  • 8) 0.000 000 000 022 599 249 92 × 2 = 0 + 0.000 000 000 045 198 499 84;
  • 9) 0.000 000 000 045 198 499 84 × 2 = 0 + 0.000 000 000 090 396 999 68;
  • 10) 0.000 000 000 090 396 999 68 × 2 = 0 + 0.000 000 000 180 793 999 36;
  • 11) 0.000 000 000 180 793 999 36 × 2 = 0 + 0.000 000 000 361 587 998 72;
  • 12) 0.000 000 000 361 587 998 72 × 2 = 0 + 0.000 000 000 723 175 997 44;
  • 13) 0.000 000 000 723 175 997 44 × 2 = 0 + 0.000 000 001 446 351 994 88;
  • 14) 0.000 000 001 446 351 994 88 × 2 = 0 + 0.000 000 002 892 703 989 76;
  • 15) 0.000 000 002 892 703 989 76 × 2 = 0 + 0.000 000 005 785 407 979 52;
  • 16) 0.000 000 005 785 407 979 52 × 2 = 0 + 0.000 000 011 570 815 959 04;
  • 17) 0.000 000 011 570 815 959 04 × 2 = 0 + 0.000 000 023 141 631 918 08;
  • 18) 0.000 000 023 141 631 918 08 × 2 = 0 + 0.000 000 046 283 263 836 16;
  • 19) 0.000 000 046 283 263 836 16 × 2 = 0 + 0.000 000 092 566 527 672 32;
  • 20) 0.000 000 092 566 527 672 32 × 2 = 0 + 0.000 000 185 133 055 344 64;
  • 21) 0.000 000 185 133 055 344 64 × 2 = 0 + 0.000 000 370 266 110 689 28;
  • 22) 0.000 000 370 266 110 689 28 × 2 = 0 + 0.000 000 740 532 221 378 56;
  • 23) 0.000 000 740 532 221 378 56 × 2 = 0 + 0.000 001 481 064 442 757 12;
  • 24) 0.000 001 481 064 442 757 12 × 2 = 0 + 0.000 002 962 128 885 514 24;
  • 25) 0.000 002 962 128 885 514 24 × 2 = 0 + 0.000 005 924 257 771 028 48;
  • 26) 0.000 005 924 257 771 028 48 × 2 = 0 + 0.000 011 848 515 542 056 96;
  • 27) 0.000 011 848 515 542 056 96 × 2 = 0 + 0.000 023 697 031 084 113 92;
  • 28) 0.000 023 697 031 084 113 92 × 2 = 0 + 0.000 047 394 062 168 227 84;
  • 29) 0.000 047 394 062 168 227 84 × 2 = 0 + 0.000 094 788 124 336 455 68;
  • 30) 0.000 094 788 124 336 455 68 × 2 = 0 + 0.000 189 576 248 672 911 36;
  • 31) 0.000 189 576 248 672 911 36 × 2 = 0 + 0.000 379 152 497 345 822 72;
  • 32) 0.000 379 152 497 345 822 72 × 2 = 0 + 0.000 758 304 994 691 645 44;
  • 33) 0.000 758 304 994 691 645 44 × 2 = 0 + 0.001 516 609 989 383 290 88;
  • 34) 0.001 516 609 989 383 290 88 × 2 = 0 + 0.003 033 219 978 766 581 76;
  • 35) 0.003 033 219 978 766 581 76 × 2 = 0 + 0.006 066 439 957 533 163 52;
  • 36) 0.006 066 439 957 533 163 52 × 2 = 0 + 0.012 132 879 915 066 327 04;
  • 37) 0.012 132 879 915 066 327 04 × 2 = 0 + 0.024 265 759 830 132 654 08;
  • 38) 0.024 265 759 830 132 654 08 × 2 = 0 + 0.048 531 519 660 265 308 16;
  • 39) 0.048 531 519 660 265 308 16 × 2 = 0 + 0.097 063 039 320 530 616 32;
  • 40) 0.097 063 039 320 530 616 32 × 2 = 0 + 0.194 126 078 641 061 232 64;
  • 41) 0.194 126 078 641 061 232 64 × 2 = 0 + 0.388 252 157 282 122 465 28;
  • 42) 0.388 252 157 282 122 465 28 × 2 = 0 + 0.776 504 314 564 244 930 56;
  • 43) 0.776 504 314 564 244 930 56 × 2 = 1 + 0.553 008 629 128 489 861 12;
  • 44) 0.553 008 629 128 489 861 12 × 2 = 1 + 0.106 017 258 256 979 722 24;
  • 45) 0.106 017 258 256 979 722 24 × 2 = 0 + 0.212 034 516 513 959 444 48;
  • 46) 0.212 034 516 513 959 444 48 × 2 = 0 + 0.424 069 033 027 918 888 96;
  • 47) 0.424 069 033 027 918 888 96 × 2 = 0 + 0.848 138 066 055 837 777 92;
  • 48) 0.848 138 066 055 837 777 92 × 2 = 1 + 0.696 276 132 111 675 555 84;
  • 49) 0.696 276 132 111 675 555 84 × 2 = 1 + 0.392 552 264 223 351 111 68;
  • 50) 0.392 552 264 223 351 111 68 × 2 = 0 + 0.785 104 528 446 702 223 36;
  • 51) 0.785 104 528 446 702 223 36 × 2 = 1 + 0.570 209 056 893 404 446 72;
  • 52) 0.570 209 056 893 404 446 72 × 2 = 1 + 0.140 418 113 786 808 893 44;
  • 53) 0.140 418 113 786 808 893 44 × 2 = 0 + 0.280 836 227 573 617 786 88;
  • 54) 0.280 836 227 573 617 786 88 × 2 = 0 + 0.561 672 455 147 235 573 76;
  • 55) 0.561 672 455 147 235 573 76 × 2 = 1 + 0.123 344 910 294 471 147 52;
  • 56) 0.123 344 910 294 471 147 52 × 2 = 0 + 0.246 689 820 588 942 295 04;
  • 57) 0.246 689 820 588 942 295 04 × 2 = 0 + 0.493 379 641 177 884 590 08;
  • 58) 0.493 379 641 177 884 590 08 × 2 = 0 + 0.986 759 282 355 769 180 16;
  • 59) 0.986 759 282 355 769 180 16 × 2 = 1 + 0.973 518 564 711 538 360 32;
  • 60) 0.973 518 564 711 538 360 32 × 2 = 1 + 0.947 037 129 423 076 720 64;
  • 61) 0.947 037 129 423 076 720 64 × 2 = 1 + 0.894 074 258 846 153 441 28;
  • 62) 0.894 074 258 846 153 441 28 × 2 = 1 + 0.788 148 517 692 306 882 56;
  • 63) 0.788 148 517 692 306 882 56 × 2 = 1 + 0.576 297 035 384 613 765 12;
  • 64) 0.576 297 035 384 613 765 12 × 2 = 1 + 0.152 594 070 769 227 530 24;
  • 65) 0.152 594 070 769 227 530 24 × 2 = 0 + 0.305 188 141 538 455 060 48;
  • 66) 0.305 188 141 538 455 060 48 × 2 = 0 + 0.610 376 283 076 910 120 96;
  • 67) 0.610 376 283 076 910 120 96 × 2 = 1 + 0.220 752 566 153 820 241 92;
  • 68) 0.220 752 566 153 820 241 92 × 2 = 0 + 0.441 505 132 307 640 483 84;
  • 69) 0.441 505 132 307 640 483 84 × 2 = 0 + 0.883 010 264 615 280 967 68;
  • 70) 0.883 010 264 615 280 967 68 × 2 = 1 + 0.766 020 529 230 561 935 36;
  • 71) 0.766 020 529 230 561 935 36 × 2 = 1 + 0.532 041 058 461 123 870 72;
  • 72) 0.532 041 058 461 123 870 72 × 2 = 1 + 0.064 082 116 922 247 741 44;
  • 73) 0.064 082 116 922 247 741 44 × 2 = 0 + 0.128 164 233 844 495 482 88;
  • 74) 0.128 164 233 844 495 482 88 × 2 = 0 + 0.256 328 467 688 990 965 76;
  • 75) 0.256 328 467 688 990 965 76 × 2 = 0 + 0.512 656 935 377 981 931 52;
  • 76) 0.512 656 935 377 981 931 52 × 2 = 1 + 0.025 313 870 755 963 863 04;
  • 77) 0.025 313 870 755 963 863 04 × 2 = 0 + 0.050 627 741 511 927 726 08;
  • 78) 0.050 627 741 511 927 726 08 × 2 = 0 + 0.101 255 483 023 855 452 16;
  • 79) 0.101 255 483 023 855 452 16 × 2 = 0 + 0.202 510 966 047 710 904 32;
  • 80) 0.202 510 966 047 710 904 32 × 2 = 0 + 0.405 021 932 095 421 808 64;
  • 81) 0.405 021 932 095 421 808 64 × 2 = 0 + 0.810 043 864 190 843 617 28;
  • 82) 0.810 043 864 190 843 617 28 × 2 = 1 + 0.620 087 728 381 687 234 56;
  • 83) 0.620 087 728 381 687 234 56 × 2 = 1 + 0.240 175 456 763 374 469 12;
  • 84) 0.240 175 456 763 374 469 12 × 2 = 0 + 0.480 350 913 526 748 938 24;
  • 85) 0.480 350 913 526 748 938 24 × 2 = 0 + 0.960 701 827 053 497 876 48;
  • 86) 0.960 701 827 053 497 876 48 × 2 = 1 + 0.921 403 654 106 995 752 96;
  • 87) 0.921 403 654 106 995 752 96 × 2 = 1 + 0.842 807 308 213 991 505 92;
  • 88) 0.842 807 308 213 991 505 92 × 2 = 1 + 0.685 614 616 427 983 011 84;
  • 89) 0.685 614 616 427 983 011 84 × 2 = 1 + 0.371 229 232 855 966 023 68;
  • 90) 0.371 229 232 855 966 023 68 × 2 = 0 + 0.742 458 465 711 932 047 36;
  • 91) 0.742 458 465 711 932 047 36 × 2 = 1 + 0.484 916 931 423 864 094 72;
  • 92) 0.484 916 931 423 864 094 72 × 2 = 0 + 0.969 833 862 847 728 189 44;
  • 93) 0.969 833 862 847 728 189 44 × 2 = 1 + 0.939 667 725 695 456 378 88;
  • 94) 0.939 667 725 695 456 378 88 × 2 = 1 + 0.879 335 451 390 912 757 76;
  • 95) 0.879 335 451 390 912 757 76 × 2 = 1 + 0.758 670 902 781 825 515 52;

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:


0.000 000 000 000 176 556 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 0010 0111 0001 0000 0110 0111 1010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 0010 0111 0001 0000 0110 0111 1010 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:


0.000 000 000 000 176 556 64(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 0010 0111 0001 0000 0110 0111 1010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1111 0010 0111 0001 0000 0110 0111 1010 111(2) × 20 =

1.1000 1101 1001 0001 1111 1001 0011 1000 1000 0011 0011 1101 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 0001 1111 1001 0011 1000 1000 0011 0011 1101 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 0001 1111 1001 0011 1000 1000 0011 0011 1101 0111 =

1000 1101 1001 0001 1111 1001 0011 1000 1000 0011 0011 1101 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 0001 1111 1001 0011 1000 1000 0011 0011 1101 0111


Decimal number -0.000 000 000 000 176 556 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1111 1001 0011 1000 1000 0011 0011 1101 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