-0.000 000 000 000 176 555 45 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 555 45(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 45(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 45| = 0.000 000 000 000 176 555 45


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 45.

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 45 × 2 = 0 + 0.000 000 000 000 353 110 9;
  • 2) 0.000 000 000 000 353 110 9 × 2 = 0 + 0.000 000 000 000 706 221 8;
  • 3) 0.000 000 000 000 706 221 8 × 2 = 0 + 0.000 000 000 001 412 443 6;
  • 4) 0.000 000 000 001 412 443 6 × 2 = 0 + 0.000 000 000 002 824 887 2;
  • 5) 0.000 000 000 002 824 887 2 × 2 = 0 + 0.000 000 000 005 649 774 4;
  • 6) 0.000 000 000 005 649 774 4 × 2 = 0 + 0.000 000 000 011 299 548 8;
  • 7) 0.000 000 000 011 299 548 8 × 2 = 0 + 0.000 000 000 022 599 097 6;
  • 8) 0.000 000 000 022 599 097 6 × 2 = 0 + 0.000 000 000 045 198 195 2;
  • 9) 0.000 000 000 045 198 195 2 × 2 = 0 + 0.000 000 000 090 396 390 4;
  • 10) 0.000 000 000 090 396 390 4 × 2 = 0 + 0.000 000 000 180 792 780 8;
  • 11) 0.000 000 000 180 792 780 8 × 2 = 0 + 0.000 000 000 361 585 561 6;
  • 12) 0.000 000 000 361 585 561 6 × 2 = 0 + 0.000 000 000 723 171 123 2;
  • 13) 0.000 000 000 723 171 123 2 × 2 = 0 + 0.000 000 001 446 342 246 4;
  • 14) 0.000 000 001 446 342 246 4 × 2 = 0 + 0.000 000 002 892 684 492 8;
  • 15) 0.000 000 002 892 684 492 8 × 2 = 0 + 0.000 000 005 785 368 985 6;
  • 16) 0.000 000 005 785 368 985 6 × 2 = 0 + 0.000 000 011 570 737 971 2;
  • 17) 0.000 000 011 570 737 971 2 × 2 = 0 + 0.000 000 023 141 475 942 4;
  • 18) 0.000 000 023 141 475 942 4 × 2 = 0 + 0.000 000 046 282 951 884 8;
  • 19) 0.000 000 046 282 951 884 8 × 2 = 0 + 0.000 000 092 565 903 769 6;
  • 20) 0.000 000 092 565 903 769 6 × 2 = 0 + 0.000 000 185 131 807 539 2;
  • 21) 0.000 000 185 131 807 539 2 × 2 = 0 + 0.000 000 370 263 615 078 4;
  • 22) 0.000 000 370 263 615 078 4 × 2 = 0 + 0.000 000 740 527 230 156 8;
  • 23) 0.000 000 740 527 230 156 8 × 2 = 0 + 0.000 001 481 054 460 313 6;
  • 24) 0.000 001 481 054 460 313 6 × 2 = 0 + 0.000 002 962 108 920 627 2;
  • 25) 0.000 002 962 108 920 627 2 × 2 = 0 + 0.000 005 924 217 841 254 4;
  • 26) 0.000 005 924 217 841 254 4 × 2 = 0 + 0.000 011 848 435 682 508 8;
  • 27) 0.000 011 848 435 682 508 8 × 2 = 0 + 0.000 023 696 871 365 017 6;
  • 28) 0.000 023 696 871 365 017 6 × 2 = 0 + 0.000 047 393 742 730 035 2;
  • 29) 0.000 047 393 742 730 035 2 × 2 = 0 + 0.000 094 787 485 460 070 4;
  • 30) 0.000 094 787 485 460 070 4 × 2 = 0 + 0.000 189 574 970 920 140 8;
  • 31) 0.000 189 574 970 920 140 8 × 2 = 0 + 0.000 379 149 941 840 281 6;
  • 32) 0.000 379 149 941 840 281 6 × 2 = 0 + 0.000 758 299 883 680 563 2;
  • 33) 0.000 758 299 883 680 563 2 × 2 = 0 + 0.001 516 599 767 361 126 4;
  • 34) 0.001 516 599 767 361 126 4 × 2 = 0 + 0.003 033 199 534 722 252 8;
  • 35) 0.003 033 199 534 722 252 8 × 2 = 0 + 0.006 066 399 069 444 505 6;
  • 36) 0.006 066 399 069 444 505 6 × 2 = 0 + 0.012 132 798 138 889 011 2;
  • 37) 0.012 132 798 138 889 011 2 × 2 = 0 + 0.024 265 596 277 778 022 4;
  • 38) 0.024 265 596 277 778 022 4 × 2 = 0 + 0.048 531 192 555 556 044 8;
  • 39) 0.048 531 192 555 556 044 8 × 2 = 0 + 0.097 062 385 111 112 089 6;
  • 40) 0.097 062 385 111 112 089 6 × 2 = 0 + 0.194 124 770 222 224 179 2;
  • 41) 0.194 124 770 222 224 179 2 × 2 = 0 + 0.388 249 540 444 448 358 4;
  • 42) 0.388 249 540 444 448 358 4 × 2 = 0 + 0.776 499 080 888 896 716 8;
  • 43) 0.776 499 080 888 896 716 8 × 2 = 1 + 0.552 998 161 777 793 433 6;
  • 44) 0.552 998 161 777 793 433 6 × 2 = 1 + 0.105 996 323 555 586 867 2;
  • 45) 0.105 996 323 555 586 867 2 × 2 = 0 + 0.211 992 647 111 173 734 4;
  • 46) 0.211 992 647 111 173 734 4 × 2 = 0 + 0.423 985 294 222 347 468 8;
  • 47) 0.423 985 294 222 347 468 8 × 2 = 0 + 0.847 970 588 444 694 937 6;
  • 48) 0.847 970 588 444 694 937 6 × 2 = 1 + 0.695 941 176 889 389 875 2;
  • 49) 0.695 941 176 889 389 875 2 × 2 = 1 + 0.391 882 353 778 779 750 4;
  • 50) 0.391 882 353 778 779 750 4 × 2 = 0 + 0.783 764 707 557 559 500 8;
  • 51) 0.783 764 707 557 559 500 8 × 2 = 1 + 0.567 529 415 115 119 001 6;
  • 52) 0.567 529 415 115 119 001 6 × 2 = 1 + 0.135 058 830 230 238 003 2;
  • 53) 0.135 058 830 230 238 003 2 × 2 = 0 + 0.270 117 660 460 476 006 4;
  • 54) 0.270 117 660 460 476 006 4 × 2 = 0 + 0.540 235 320 920 952 012 8;
  • 55) 0.540 235 320 920 952 012 8 × 2 = 1 + 0.080 470 641 841 904 025 6;
  • 56) 0.080 470 641 841 904 025 6 × 2 = 0 + 0.160 941 283 683 808 051 2;
  • 57) 0.160 941 283 683 808 051 2 × 2 = 0 + 0.321 882 567 367 616 102 4;
  • 58) 0.321 882 567 367 616 102 4 × 2 = 0 + 0.643 765 134 735 232 204 8;
  • 59) 0.643 765 134 735 232 204 8 × 2 = 1 + 0.287 530 269 470 464 409 6;
  • 60) 0.287 530 269 470 464 409 6 × 2 = 0 + 0.575 060 538 940 928 819 2;
  • 61) 0.575 060 538 940 928 819 2 × 2 = 1 + 0.150 121 077 881 857 638 4;
  • 62) 0.150 121 077 881 857 638 4 × 2 = 0 + 0.300 242 155 763 715 276 8;
  • 63) 0.300 242 155 763 715 276 8 × 2 = 0 + 0.600 484 311 527 430 553 6;
  • 64) 0.600 484 311 527 430 553 6 × 2 = 1 + 0.200 968 623 054 861 107 2;
  • 65) 0.200 968 623 054 861 107 2 × 2 = 0 + 0.401 937 246 109 722 214 4;
  • 66) 0.401 937 246 109 722 214 4 × 2 = 0 + 0.803 874 492 219 444 428 8;
  • 67) 0.803 874 492 219 444 428 8 × 2 = 1 + 0.607 748 984 438 888 857 6;
  • 68) 0.607 748 984 438 888 857 6 × 2 = 1 + 0.215 497 968 877 777 715 2;
  • 69) 0.215 497 968 877 777 715 2 × 2 = 0 + 0.430 995 937 755 555 430 4;
  • 70) 0.430 995 937 755 555 430 4 × 2 = 0 + 0.861 991 875 511 110 860 8;
  • 71) 0.861 991 875 511 110 860 8 × 2 = 1 + 0.723 983 751 022 221 721 6;
  • 72) 0.723 983 751 022 221 721 6 × 2 = 1 + 0.447 967 502 044 443 443 2;
  • 73) 0.447 967 502 044 443 443 2 × 2 = 0 + 0.895 935 004 088 886 886 4;
  • 74) 0.895 935 004 088 886 886 4 × 2 = 1 + 0.791 870 008 177 773 772 8;
  • 75) 0.791 870 008 177 773 772 8 × 2 = 1 + 0.583 740 016 355 547 545 6;
  • 76) 0.583 740 016 355 547 545 6 × 2 = 1 + 0.167 480 032 711 095 091 2;
  • 77) 0.167 480 032 711 095 091 2 × 2 = 0 + 0.334 960 065 422 190 182 4;
  • 78) 0.334 960 065 422 190 182 4 × 2 = 0 + 0.669 920 130 844 380 364 8;
  • 79) 0.669 920 130 844 380 364 8 × 2 = 1 + 0.339 840 261 688 760 729 6;
  • 80) 0.339 840 261 688 760 729 6 × 2 = 0 + 0.679 680 523 377 521 459 2;
  • 81) 0.679 680 523 377 521 459 2 × 2 = 1 + 0.359 361 046 755 042 918 4;
  • 82) 0.359 361 046 755 042 918 4 × 2 = 0 + 0.718 722 093 510 085 836 8;
  • 83) 0.718 722 093 510 085 836 8 × 2 = 1 + 0.437 444 187 020 171 673 6;
  • 84) 0.437 444 187 020 171 673 6 × 2 = 0 + 0.874 888 374 040 343 347 2;
  • 85) 0.874 888 374 040 343 347 2 × 2 = 1 + 0.749 776 748 080 686 694 4;
  • 86) 0.749 776 748 080 686 694 4 × 2 = 1 + 0.499 553 496 161 373 388 8;
  • 87) 0.499 553 496 161 373 388 8 × 2 = 0 + 0.999 106 992 322 746 777 6;
  • 88) 0.999 106 992 322 746 777 6 × 2 = 1 + 0.998 213 984 645 493 555 2;
  • 89) 0.998 213 984 645 493 555 2 × 2 = 1 + 0.996 427 969 290 987 110 4;
  • 90) 0.996 427 969 290 987 110 4 × 2 = 1 + 0.992 855 938 581 974 220 8;
  • 91) 0.992 855 938 581 974 220 8 × 2 = 1 + 0.985 711 877 163 948 441 6;
  • 92) 0.985 711 877 163 948 441 6 × 2 = 1 + 0.971 423 754 327 896 883 2;
  • 93) 0.971 423 754 327 896 883 2 × 2 = 1 + 0.942 847 508 655 793 766 4;
  • 94) 0.942 847 508 655 793 766 4 × 2 = 1 + 0.885 695 017 311 587 532 8;
  • 95) 0.885 695 017 311 587 532 8 × 2 = 1 + 0.771 390 034 623 175 065 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:


0.000 000 000 000 176 555 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1001 0011 0011 0111 0010 1010 1101 1111 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 555 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1001 0011 0011 0111 0010 1010 1101 1111 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 45(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1001 0011 0011 0111 0010 1010 1101 1111 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0010 1001 0011 0011 0111 0010 1010 1101 1111 111(2) × 20 =

1.1000 1101 1001 0001 0100 1001 1001 1011 1001 0101 0110 1111 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 0100 1001 1001 1011 1001 0101 0110 1111 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 0100 1001 1001 1011 1001 0101 0110 1111 1111 =

1000 1101 1001 0001 0100 1001 1001 1011 1001 0101 0110 1111 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 0100 1001 1001 1011 1001 0101 0110 1111 1111


Decimal number -0.000 000 000 000 176 555 45 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 0100 1001 1001 1011 1001 0101 0110 1111 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