-0.000 000 000 000 176 556 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 19(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 19(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 19| = 0.000 000 000 000 176 556 19


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 19.

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 19 × 2 = 0 + 0.000 000 000 000 353 112 38;
  • 2) 0.000 000 000 000 353 112 38 × 2 = 0 + 0.000 000 000 000 706 224 76;
  • 3) 0.000 000 000 000 706 224 76 × 2 = 0 + 0.000 000 000 001 412 449 52;
  • 4) 0.000 000 000 001 412 449 52 × 2 = 0 + 0.000 000 000 002 824 899 04;
  • 5) 0.000 000 000 002 824 899 04 × 2 = 0 + 0.000 000 000 005 649 798 08;
  • 6) 0.000 000 000 005 649 798 08 × 2 = 0 + 0.000 000 000 011 299 596 16;
  • 7) 0.000 000 000 011 299 596 16 × 2 = 0 + 0.000 000 000 022 599 192 32;
  • 8) 0.000 000 000 022 599 192 32 × 2 = 0 + 0.000 000 000 045 198 384 64;
  • 9) 0.000 000 000 045 198 384 64 × 2 = 0 + 0.000 000 000 090 396 769 28;
  • 10) 0.000 000 000 090 396 769 28 × 2 = 0 + 0.000 000 000 180 793 538 56;
  • 11) 0.000 000 000 180 793 538 56 × 2 = 0 + 0.000 000 000 361 587 077 12;
  • 12) 0.000 000 000 361 587 077 12 × 2 = 0 + 0.000 000 000 723 174 154 24;
  • 13) 0.000 000 000 723 174 154 24 × 2 = 0 + 0.000 000 001 446 348 308 48;
  • 14) 0.000 000 001 446 348 308 48 × 2 = 0 + 0.000 000 002 892 696 616 96;
  • 15) 0.000 000 002 892 696 616 96 × 2 = 0 + 0.000 000 005 785 393 233 92;
  • 16) 0.000 000 005 785 393 233 92 × 2 = 0 + 0.000 000 011 570 786 467 84;
  • 17) 0.000 000 011 570 786 467 84 × 2 = 0 + 0.000 000 023 141 572 935 68;
  • 18) 0.000 000 023 141 572 935 68 × 2 = 0 + 0.000 000 046 283 145 871 36;
  • 19) 0.000 000 046 283 145 871 36 × 2 = 0 + 0.000 000 092 566 291 742 72;
  • 20) 0.000 000 092 566 291 742 72 × 2 = 0 + 0.000 000 185 132 583 485 44;
  • 21) 0.000 000 185 132 583 485 44 × 2 = 0 + 0.000 000 370 265 166 970 88;
  • 22) 0.000 000 370 265 166 970 88 × 2 = 0 + 0.000 000 740 530 333 941 76;
  • 23) 0.000 000 740 530 333 941 76 × 2 = 0 + 0.000 001 481 060 667 883 52;
  • 24) 0.000 001 481 060 667 883 52 × 2 = 0 + 0.000 002 962 121 335 767 04;
  • 25) 0.000 002 962 121 335 767 04 × 2 = 0 + 0.000 005 924 242 671 534 08;
  • 26) 0.000 005 924 242 671 534 08 × 2 = 0 + 0.000 011 848 485 343 068 16;
  • 27) 0.000 011 848 485 343 068 16 × 2 = 0 + 0.000 023 696 970 686 136 32;
  • 28) 0.000 023 696 970 686 136 32 × 2 = 0 + 0.000 047 393 941 372 272 64;
  • 29) 0.000 047 393 941 372 272 64 × 2 = 0 + 0.000 094 787 882 744 545 28;
  • 30) 0.000 094 787 882 744 545 28 × 2 = 0 + 0.000 189 575 765 489 090 56;
  • 31) 0.000 189 575 765 489 090 56 × 2 = 0 + 0.000 379 151 530 978 181 12;
  • 32) 0.000 379 151 530 978 181 12 × 2 = 0 + 0.000 758 303 061 956 362 24;
  • 33) 0.000 758 303 061 956 362 24 × 2 = 0 + 0.001 516 606 123 912 724 48;
  • 34) 0.001 516 606 123 912 724 48 × 2 = 0 + 0.003 033 212 247 825 448 96;
  • 35) 0.003 033 212 247 825 448 96 × 2 = 0 + 0.006 066 424 495 650 897 92;
  • 36) 0.006 066 424 495 650 897 92 × 2 = 0 + 0.012 132 848 991 301 795 84;
  • 37) 0.012 132 848 991 301 795 84 × 2 = 0 + 0.024 265 697 982 603 591 68;
  • 38) 0.024 265 697 982 603 591 68 × 2 = 0 + 0.048 531 395 965 207 183 36;
  • 39) 0.048 531 395 965 207 183 36 × 2 = 0 + 0.097 062 791 930 414 366 72;
  • 40) 0.097 062 791 930 414 366 72 × 2 = 0 + 0.194 125 583 860 828 733 44;
  • 41) 0.194 125 583 860 828 733 44 × 2 = 0 + 0.388 251 167 721 657 466 88;
  • 42) 0.388 251 167 721 657 466 88 × 2 = 0 + 0.776 502 335 443 314 933 76;
  • 43) 0.776 502 335 443 314 933 76 × 2 = 1 + 0.553 004 670 886 629 867 52;
  • 44) 0.553 004 670 886 629 867 52 × 2 = 1 + 0.106 009 341 773 259 735 04;
  • 45) 0.106 009 341 773 259 735 04 × 2 = 0 + 0.212 018 683 546 519 470 08;
  • 46) 0.212 018 683 546 519 470 08 × 2 = 0 + 0.424 037 367 093 038 940 16;
  • 47) 0.424 037 367 093 038 940 16 × 2 = 0 + 0.848 074 734 186 077 880 32;
  • 48) 0.848 074 734 186 077 880 32 × 2 = 1 + 0.696 149 468 372 155 760 64;
  • 49) 0.696 149 468 372 155 760 64 × 2 = 1 + 0.392 298 936 744 311 521 28;
  • 50) 0.392 298 936 744 311 521 28 × 2 = 0 + 0.784 597 873 488 623 042 56;
  • 51) 0.784 597 873 488 623 042 56 × 2 = 1 + 0.569 195 746 977 246 085 12;
  • 52) 0.569 195 746 977 246 085 12 × 2 = 1 + 0.138 391 493 954 492 170 24;
  • 53) 0.138 391 493 954 492 170 24 × 2 = 0 + 0.276 782 987 908 984 340 48;
  • 54) 0.276 782 987 908 984 340 48 × 2 = 0 + 0.553 565 975 817 968 680 96;
  • 55) 0.553 565 975 817 968 680 96 × 2 = 1 + 0.107 131 951 635 937 361 92;
  • 56) 0.107 131 951 635 937 361 92 × 2 = 0 + 0.214 263 903 271 874 723 84;
  • 57) 0.214 263 903 271 874 723 84 × 2 = 0 + 0.428 527 806 543 749 447 68;
  • 58) 0.428 527 806 543 749 447 68 × 2 = 0 + 0.857 055 613 087 498 895 36;
  • 59) 0.857 055 613 087 498 895 36 × 2 = 1 + 0.714 111 226 174 997 790 72;
  • 60) 0.714 111 226 174 997 790 72 × 2 = 1 + 0.428 222 452 349 995 581 44;
  • 61) 0.428 222 452 349 995 581 44 × 2 = 0 + 0.856 444 904 699 991 162 88;
  • 62) 0.856 444 904 699 991 162 88 × 2 = 1 + 0.712 889 809 399 982 325 76;
  • 63) 0.712 889 809 399 982 325 76 × 2 = 1 + 0.425 779 618 799 964 651 52;
  • 64) 0.425 779 618 799 964 651 52 × 2 = 0 + 0.851 559 237 599 929 303 04;
  • 65) 0.851 559 237 599 929 303 04 × 2 = 1 + 0.703 118 475 199 858 606 08;
  • 66) 0.703 118 475 199 858 606 08 × 2 = 1 + 0.406 236 950 399 717 212 16;
  • 67) 0.406 236 950 399 717 212 16 × 2 = 0 + 0.812 473 900 799 434 424 32;
  • 68) 0.812 473 900 799 434 424 32 × 2 = 1 + 0.624 947 801 598 868 848 64;
  • 69) 0.624 947 801 598 868 848 64 × 2 = 1 + 0.249 895 603 197 737 697 28;
  • 70) 0.249 895 603 197 737 697 28 × 2 = 0 + 0.499 791 206 395 475 394 56;
  • 71) 0.499 791 206 395 475 394 56 × 2 = 0 + 0.999 582 412 790 950 789 12;
  • 72) 0.999 582 412 790 950 789 12 × 2 = 1 + 0.999 164 825 581 901 578 24;
  • 73) 0.999 164 825 581 901 578 24 × 2 = 1 + 0.998 329 651 163 803 156 48;
  • 74) 0.998 329 651 163 803 156 48 × 2 = 1 + 0.996 659 302 327 606 312 96;
  • 75) 0.996 659 302 327 606 312 96 × 2 = 1 + 0.993 318 604 655 212 625 92;
  • 76) 0.993 318 604 655 212 625 92 × 2 = 1 + 0.986 637 209 310 425 251 84;
  • 77) 0.986 637 209 310 425 251 84 × 2 = 1 + 0.973 274 418 620 850 503 68;
  • 78) 0.973 274 418 620 850 503 68 × 2 = 1 + 0.946 548 837 241 701 007 36;
  • 79) 0.946 548 837 241 701 007 36 × 2 = 1 + 0.893 097 674 483 402 014 72;
  • 80) 0.893 097 674 483 402 014 72 × 2 = 1 + 0.786 195 348 966 804 029 44;
  • 81) 0.786 195 348 966 804 029 44 × 2 = 1 + 0.572 390 697 933 608 058 88;
  • 82) 0.572 390 697 933 608 058 88 × 2 = 1 + 0.144 781 395 867 216 117 76;
  • 83) 0.144 781 395 867 216 117 76 × 2 = 0 + 0.289 562 791 734 432 235 52;
  • 84) 0.289 562 791 734 432 235 52 × 2 = 0 + 0.579 125 583 468 864 471 04;
  • 85) 0.579 125 583 468 864 471 04 × 2 = 1 + 0.158 251 166 937 728 942 08;
  • 86) 0.158 251 166 937 728 942 08 × 2 = 0 + 0.316 502 333 875 457 884 16;
  • 87) 0.316 502 333 875 457 884 16 × 2 = 0 + 0.633 004 667 750 915 768 32;
  • 88) 0.633 004 667 750 915 768 32 × 2 = 1 + 0.266 009 335 501 831 536 64;
  • 89) 0.266 009 335 501 831 536 64 × 2 = 0 + 0.532 018 671 003 663 073 28;
  • 90) 0.532 018 671 003 663 073 28 × 2 = 1 + 0.064 037 342 007 326 146 56;
  • 91) 0.064 037 342 007 326 146 56 × 2 = 0 + 0.128 074 684 014 652 293 12;
  • 92) 0.128 074 684 014 652 293 12 × 2 = 0 + 0.256 149 368 029 304 586 24;
  • 93) 0.256 149 368 029 304 586 24 × 2 = 0 + 0.512 298 736 058 609 172 48;
  • 94) 0.512 298 736 058 609 172 48 × 2 = 1 + 0.024 597 472 117 218 344 96;
  • 95) 0.024 597 472 117 218 344 96 × 2 = 0 + 0.049 194 944 234 436 689 92;

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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0110 1101 1001 1111 1111 1100 1001 0100 010(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0110 1101 1001 1111 1111 1100 1001 0100 010(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 19(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0110 1101 1001 1111 1111 1100 1001 0100 010(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 0110 1101 1001 1111 1111 1100 1001 0100 010(2) × 20 =

1.1000 1101 1001 0001 1011 0110 1100 1111 1111 1110 0100 1010 0010(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 1011 0110 1100 1111 1111 1110 0100 1010 0010


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 1011 0110 1100 1111 1111 1110 0100 1010 0010 =

1000 1101 1001 0001 1011 0110 1100 1111 1111 1110 0100 1010 0010


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 1011 0110 1100 1111 1111 1110 0100 1010 0010


Decimal number -0.000 000 000 000 176 556 19 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1011 0110 1100 1111 1111 1110 0100 1010 0010

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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