-0.000 000 000 000 176 558 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 558 26(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
-0.000 000 000 000 176 558 26(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:

|-0.000 000 000 000 176 558 26| = 0.000 000 000 000 176 558 26


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: 0.000 000 000 000 176 558 26.

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) 0.000 000 000 000 176 558 26 × 2 = 0 + 0.000 000 000 000 353 116 52;
  • 2) 0.000 000 000 000 353 116 52 × 2 = 0 + 0.000 000 000 000 706 233 04;
  • 3) 0.000 000 000 000 706 233 04 × 2 = 0 + 0.000 000 000 001 412 466 08;
  • 4) 0.000 000 000 001 412 466 08 × 2 = 0 + 0.000 000 000 002 824 932 16;
  • 5) 0.000 000 000 002 824 932 16 × 2 = 0 + 0.000 000 000 005 649 864 32;
  • 6) 0.000 000 000 005 649 864 32 × 2 = 0 + 0.000 000 000 011 299 728 64;
  • 7) 0.000 000 000 011 299 728 64 × 2 = 0 + 0.000 000 000 022 599 457 28;
  • 8) 0.000 000 000 022 599 457 28 × 2 = 0 + 0.000 000 000 045 198 914 56;
  • 9) 0.000 000 000 045 198 914 56 × 2 = 0 + 0.000 000 000 090 397 829 12;
  • 10) 0.000 000 000 090 397 829 12 × 2 = 0 + 0.000 000 000 180 795 658 24;
  • 11) 0.000 000 000 180 795 658 24 × 2 = 0 + 0.000 000 000 361 591 316 48;
  • 12) 0.000 000 000 361 591 316 48 × 2 = 0 + 0.000 000 000 723 182 632 96;
  • 13) 0.000 000 000 723 182 632 96 × 2 = 0 + 0.000 000 001 446 365 265 92;
  • 14) 0.000 000 001 446 365 265 92 × 2 = 0 + 0.000 000 002 892 730 531 84;
  • 15) 0.000 000 002 892 730 531 84 × 2 = 0 + 0.000 000 005 785 461 063 68;
  • 16) 0.000 000 005 785 461 063 68 × 2 = 0 + 0.000 000 011 570 922 127 36;
  • 17) 0.000 000 011 570 922 127 36 × 2 = 0 + 0.000 000 023 141 844 254 72;
  • 18) 0.000 000 023 141 844 254 72 × 2 = 0 + 0.000 000 046 283 688 509 44;
  • 19) 0.000 000 046 283 688 509 44 × 2 = 0 + 0.000 000 092 567 377 018 88;
  • 20) 0.000 000 092 567 377 018 88 × 2 = 0 + 0.000 000 185 134 754 037 76;
  • 21) 0.000 000 185 134 754 037 76 × 2 = 0 + 0.000 000 370 269 508 075 52;
  • 22) 0.000 000 370 269 508 075 52 × 2 = 0 + 0.000 000 740 539 016 151 04;
  • 23) 0.000 000 740 539 016 151 04 × 2 = 0 + 0.000 001 481 078 032 302 08;
  • 24) 0.000 001 481 078 032 302 08 × 2 = 0 + 0.000 002 962 156 064 604 16;
  • 25) 0.000 002 962 156 064 604 16 × 2 = 0 + 0.000 005 924 312 129 208 32;
  • 26) 0.000 005 924 312 129 208 32 × 2 = 0 + 0.000 011 848 624 258 416 64;
  • 27) 0.000 011 848 624 258 416 64 × 2 = 0 + 0.000 023 697 248 516 833 28;
  • 28) 0.000 023 697 248 516 833 28 × 2 = 0 + 0.000 047 394 497 033 666 56;
  • 29) 0.000 047 394 497 033 666 56 × 2 = 0 + 0.000 094 788 994 067 333 12;
  • 30) 0.000 094 788 994 067 333 12 × 2 = 0 + 0.000 189 577 988 134 666 24;
  • 31) 0.000 189 577 988 134 666 24 × 2 = 0 + 0.000 379 155 976 269 332 48;
  • 32) 0.000 379 155 976 269 332 48 × 2 = 0 + 0.000 758 311 952 538 664 96;
  • 33) 0.000 758 311 952 538 664 96 × 2 = 0 + 0.001 516 623 905 077 329 92;
  • 34) 0.001 516 623 905 077 329 92 × 2 = 0 + 0.003 033 247 810 154 659 84;
  • 35) 0.003 033 247 810 154 659 84 × 2 = 0 + 0.006 066 495 620 309 319 68;
  • 36) 0.006 066 495 620 309 319 68 × 2 = 0 + 0.012 132 991 240 618 639 36;
  • 37) 0.012 132 991 240 618 639 36 × 2 = 0 + 0.024 265 982 481 237 278 72;
  • 38) 0.024 265 982 481 237 278 72 × 2 = 0 + 0.048 531 964 962 474 557 44;
  • 39) 0.048 531 964 962 474 557 44 × 2 = 0 + 0.097 063 929 924 949 114 88;
  • 40) 0.097 063 929 924 949 114 88 × 2 = 0 + 0.194 127 859 849 898 229 76;
  • 41) 0.194 127 859 849 898 229 76 × 2 = 0 + 0.388 255 719 699 796 459 52;
  • 42) 0.388 255 719 699 796 459 52 × 2 = 0 + 0.776 511 439 399 592 919 04;
  • 43) 0.776 511 439 399 592 919 04 × 2 = 1 + 0.553 022 878 799 185 838 08;
  • 44) 0.553 022 878 799 185 838 08 × 2 = 1 + 0.106 045 757 598 371 676 16;
  • 45) 0.106 045 757 598 371 676 16 × 2 = 0 + 0.212 091 515 196 743 352 32;
  • 46) 0.212 091 515 196 743 352 32 × 2 = 0 + 0.424 183 030 393 486 704 64;
  • 47) 0.424 183 030 393 486 704 64 × 2 = 0 + 0.848 366 060 786 973 409 28;
  • 48) 0.848 366 060 786 973 409 28 × 2 = 1 + 0.696 732 121 573 946 818 56;
  • 49) 0.696 732 121 573 946 818 56 × 2 = 1 + 0.393 464 243 147 893 637 12;
  • 50) 0.393 464 243 147 893 637 12 × 2 = 0 + 0.786 928 486 295 787 274 24;
  • 51) 0.786 928 486 295 787 274 24 × 2 = 1 + 0.573 856 972 591 574 548 48;
  • 52) 0.573 856 972 591 574 548 48 × 2 = 1 + 0.147 713 945 183 149 096 96;
  • 53) 0.147 713 945 183 149 096 96 × 2 = 0 + 0.295 427 890 366 298 193 92;
  • 54) 0.295 427 890 366 298 193 92 × 2 = 0 + 0.590 855 780 732 596 387 84;
  • 55) 0.590 855 780 732 596 387 84 × 2 = 1 + 0.181 711 561 465 192 775 68;
  • 56) 0.181 711 561 465 192 775 68 × 2 = 0 + 0.363 423 122 930 385 551 36;
  • 57) 0.363 423 122 930 385 551 36 × 2 = 0 + 0.726 846 245 860 771 102 72;
  • 58) 0.726 846 245 860 771 102 72 × 2 = 1 + 0.453 692 491 721 542 205 44;
  • 59) 0.453 692 491 721 542 205 44 × 2 = 0 + 0.907 384 983 443 084 410 88;
  • 60) 0.907 384 983 443 084 410 88 × 2 = 1 + 0.814 769 966 886 168 821 76;
  • 61) 0.814 769 966 886 168 821 76 × 2 = 1 + 0.629 539 933 772 337 643 52;
  • 62) 0.629 539 933 772 337 643 52 × 2 = 1 + 0.259 079 867 544 675 287 04;
  • 63) 0.259 079 867 544 675 287 04 × 2 = 0 + 0.518 159 735 089 350 574 08;
  • 64) 0.518 159 735 089 350 574 08 × 2 = 1 + 0.036 319 470 178 701 148 16;
  • 65) 0.036 319 470 178 701 148 16 × 2 = 0 + 0.072 638 940 357 402 296 32;
  • 66) 0.072 638 940 357 402 296 32 × 2 = 0 + 0.145 277 880 714 804 592 64;
  • 67) 0.145 277 880 714 804 592 64 × 2 = 0 + 0.290 555 761 429 609 185 28;
  • 68) 0.290 555 761 429 609 185 28 × 2 = 0 + 0.581 111 522 859 218 370 56;
  • 69) 0.581 111 522 859 218 370 56 × 2 = 1 + 0.162 223 045 718 436 741 12;
  • 70) 0.162 223 045 718 436 741 12 × 2 = 0 + 0.324 446 091 436 873 482 24;
  • 71) 0.324 446 091 436 873 482 24 × 2 = 0 + 0.648 892 182 873 746 964 48;
  • 72) 0.648 892 182 873 746 964 48 × 2 = 1 + 0.297 784 365 747 493 928 96;
  • 73) 0.297 784 365 747 493 928 96 × 2 = 0 + 0.595 568 731 494 987 857 92;
  • 74) 0.595 568 731 494 987 857 92 × 2 = 1 + 0.191 137 462 989 975 715 84;
  • 75) 0.191 137 462 989 975 715 84 × 2 = 0 + 0.382 274 925 979 951 431 68;
  • 76) 0.382 274 925 979 951 431 68 × 2 = 0 + 0.764 549 851 959 902 863 36;
  • 77) 0.764 549 851 959 902 863 36 × 2 = 1 + 0.529 099 703 919 805 726 72;
  • 78) 0.529 099 703 919 805 726 72 × 2 = 1 + 0.058 199 407 839 611 453 44;
  • 79) 0.058 199 407 839 611 453 44 × 2 = 0 + 0.116 398 815 679 222 906 88;
  • 80) 0.116 398 815 679 222 906 88 × 2 = 0 + 0.232 797 631 358 445 813 76;
  • 81) 0.232 797 631 358 445 813 76 × 2 = 0 + 0.465 595 262 716 891 627 52;
  • 82) 0.465 595 262 716 891 627 52 × 2 = 0 + 0.931 190 525 433 783 255 04;
  • 83) 0.931 190 525 433 783 255 04 × 2 = 1 + 0.862 381 050 867 566 510 08;
  • 84) 0.862 381 050 867 566 510 08 × 2 = 1 + 0.724 762 101 735 133 020 16;
  • 85) 0.724 762 101 735 133 020 16 × 2 = 1 + 0.449 524 203 470 266 040 32;
  • 86) 0.449 524 203 470 266 040 32 × 2 = 0 + 0.899 048 406 940 532 080 64;
  • 87) 0.899 048 406 940 532 080 64 × 2 = 1 + 0.798 096 813 881 064 161 28;
  • 88) 0.798 096 813 881 064 161 28 × 2 = 1 + 0.596 193 627 762 128 322 56;
  • 89) 0.596 193 627 762 128 322 56 × 2 = 1 + 0.192 387 255 524 256 645 12;
  • 90) 0.192 387 255 524 256 645 12 × 2 = 0 + 0.384 774 511 048 513 290 24;
  • 91) 0.384 774 511 048 513 290 24 × 2 = 0 + 0.769 549 022 097 026 580 48;
  • 92) 0.769 549 022 097 026 580 48 × 2 = 1 + 0.539 098 044 194 053 160 96;
  • 93) 0.539 098 044 194 053 160 96 × 2 = 1 + 0.078 196 088 388 106 321 92;
  • 94) 0.078 196 088 388 106 321 92 × 2 = 0 + 0.156 392 176 776 212 643 84;
  • 95) 0.156 392 176 776 212 643 84 × 2 = 0 + 0.312 784 353 552 425 287 68;

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:


0.000 000 000 000 176 558 26(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1101 0000 1001 0100 1100 0011 1011 1001 100(2)

6. Positive number before normalization:

0.000 000 000 000 176 558 26(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1101 0000 1001 0100 1100 0011 1011 1001 100(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:


0.000 000 000 000 176 558 26(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1101 0000 1001 0100 1100 0011 1011 1001 100(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0101 1101 0000 1001 0100 1100 0011 1011 1001 100(2) × 20 =

1.1000 1101 1001 0010 1110 1000 0100 1010 0110 0001 1101 1100 1100(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 1110 1000 0100 1010 0110 0001 1101 1100 1100


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 1110 1000 0100 1010 0110 0001 1101 1100 1100 =

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


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 1110 1000 0100 1010 0110 0001 1101 1100 1100


Decimal number -0.000 000 000 000 176 558 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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