-0.000 000 000 000 176 540 4 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 540 4(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
-0.000 000 000 000 176 540 4(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:

|-0.000 000 000 000 176 540 4| = 0.000 000 000 000 176 540 4


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: 0.000 000 000 000 176 540 4.

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) 0.000 000 000 000 176 540 4 × 2 = 0 + 0.000 000 000 000 353 080 8;
  • 2) 0.000 000 000 000 353 080 8 × 2 = 0 + 0.000 000 000 000 706 161 6;
  • 3) 0.000 000 000 000 706 161 6 × 2 = 0 + 0.000 000 000 001 412 323 2;
  • 4) 0.000 000 000 001 412 323 2 × 2 = 0 + 0.000 000 000 002 824 646 4;
  • 5) 0.000 000 000 002 824 646 4 × 2 = 0 + 0.000 000 000 005 649 292 8;
  • 6) 0.000 000 000 005 649 292 8 × 2 = 0 + 0.000 000 000 011 298 585 6;
  • 7) 0.000 000 000 011 298 585 6 × 2 = 0 + 0.000 000 000 022 597 171 2;
  • 8) 0.000 000 000 022 597 171 2 × 2 = 0 + 0.000 000 000 045 194 342 4;
  • 9) 0.000 000 000 045 194 342 4 × 2 = 0 + 0.000 000 000 090 388 684 8;
  • 10) 0.000 000 000 090 388 684 8 × 2 = 0 + 0.000 000 000 180 777 369 6;
  • 11) 0.000 000 000 180 777 369 6 × 2 = 0 + 0.000 000 000 361 554 739 2;
  • 12) 0.000 000 000 361 554 739 2 × 2 = 0 + 0.000 000 000 723 109 478 4;
  • 13) 0.000 000 000 723 109 478 4 × 2 = 0 + 0.000 000 001 446 218 956 8;
  • 14) 0.000 000 001 446 218 956 8 × 2 = 0 + 0.000 000 002 892 437 913 6;
  • 15) 0.000 000 002 892 437 913 6 × 2 = 0 + 0.000 000 005 784 875 827 2;
  • 16) 0.000 000 005 784 875 827 2 × 2 = 0 + 0.000 000 011 569 751 654 4;
  • 17) 0.000 000 011 569 751 654 4 × 2 = 0 + 0.000 000 023 139 503 308 8;
  • 18) 0.000 000 023 139 503 308 8 × 2 = 0 + 0.000 000 046 279 006 617 6;
  • 19) 0.000 000 046 279 006 617 6 × 2 = 0 + 0.000 000 092 558 013 235 2;
  • 20) 0.000 000 092 558 013 235 2 × 2 = 0 + 0.000 000 185 116 026 470 4;
  • 21) 0.000 000 185 116 026 470 4 × 2 = 0 + 0.000 000 370 232 052 940 8;
  • 22) 0.000 000 370 232 052 940 8 × 2 = 0 + 0.000 000 740 464 105 881 6;
  • 23) 0.000 000 740 464 105 881 6 × 2 = 0 + 0.000 001 480 928 211 763 2;
  • 24) 0.000 001 480 928 211 763 2 × 2 = 0 + 0.000 002 961 856 423 526 4;
  • 25) 0.000 002 961 856 423 526 4 × 2 = 0 + 0.000 005 923 712 847 052 8;
  • 26) 0.000 005 923 712 847 052 8 × 2 = 0 + 0.000 011 847 425 694 105 6;
  • 27) 0.000 011 847 425 694 105 6 × 2 = 0 + 0.000 023 694 851 388 211 2;
  • 28) 0.000 023 694 851 388 211 2 × 2 = 0 + 0.000 047 389 702 776 422 4;
  • 29) 0.000 047 389 702 776 422 4 × 2 = 0 + 0.000 094 779 405 552 844 8;
  • 30) 0.000 094 779 405 552 844 8 × 2 = 0 + 0.000 189 558 811 105 689 6;
  • 31) 0.000 189 558 811 105 689 6 × 2 = 0 + 0.000 379 117 622 211 379 2;
  • 32) 0.000 379 117 622 211 379 2 × 2 = 0 + 0.000 758 235 244 422 758 4;
  • 33) 0.000 758 235 244 422 758 4 × 2 = 0 + 0.001 516 470 488 845 516 8;
  • 34) 0.001 516 470 488 845 516 8 × 2 = 0 + 0.003 032 940 977 691 033 6;
  • 35) 0.003 032 940 977 691 033 6 × 2 = 0 + 0.006 065 881 955 382 067 2;
  • 36) 0.006 065 881 955 382 067 2 × 2 = 0 + 0.012 131 763 910 764 134 4;
  • 37) 0.012 131 763 910 764 134 4 × 2 = 0 + 0.024 263 527 821 528 268 8;
  • 38) 0.024 263 527 821 528 268 8 × 2 = 0 + 0.048 527 055 643 056 537 6;
  • 39) 0.048 527 055 643 056 537 6 × 2 = 0 + 0.097 054 111 286 113 075 2;
  • 40) 0.097 054 111 286 113 075 2 × 2 = 0 + 0.194 108 222 572 226 150 4;
  • 41) 0.194 108 222 572 226 150 4 × 2 = 0 + 0.388 216 445 144 452 300 8;
  • 42) 0.388 216 445 144 452 300 8 × 2 = 0 + 0.776 432 890 288 904 601 6;
  • 43) 0.776 432 890 288 904 601 6 × 2 = 1 + 0.552 865 780 577 809 203 2;
  • 44) 0.552 865 780 577 809 203 2 × 2 = 1 + 0.105 731 561 155 618 406 4;
  • 45) 0.105 731 561 155 618 406 4 × 2 = 0 + 0.211 463 122 311 236 812 8;
  • 46) 0.211 463 122 311 236 812 8 × 2 = 0 + 0.422 926 244 622 473 625 6;
  • 47) 0.422 926 244 622 473 625 6 × 2 = 0 + 0.845 852 489 244 947 251 2;
  • 48) 0.845 852 489 244 947 251 2 × 2 = 1 + 0.691 704 978 489 894 502 4;
  • 49) 0.691 704 978 489 894 502 4 × 2 = 1 + 0.383 409 956 979 789 004 8;
  • 50) 0.383 409 956 979 789 004 8 × 2 = 0 + 0.766 819 913 959 578 009 6;
  • 51) 0.766 819 913 959 578 009 6 × 2 = 1 + 0.533 639 827 919 156 019 2;
  • 52) 0.533 639 827 919 156 019 2 × 2 = 1 + 0.067 279 655 838 312 038 4;
  • 53) 0.067 279 655 838 312 038 4 × 2 = 0 + 0.134 559 311 676 624 076 8;
  • 54) 0.134 559 311 676 624 076 8 × 2 = 0 + 0.269 118 623 353 248 153 6;
  • 55) 0.269 118 623 353 248 153 6 × 2 = 0 + 0.538 237 246 706 496 307 2;
  • 56) 0.538 237 246 706 496 307 2 × 2 = 1 + 0.076 474 493 412 992 614 4;
  • 57) 0.076 474 493 412 992 614 4 × 2 = 0 + 0.152 948 986 825 985 228 8;
  • 58) 0.152 948 986 825 985 228 8 × 2 = 0 + 0.305 897 973 651 970 457 6;
  • 59) 0.305 897 973 651 970 457 6 × 2 = 0 + 0.611 795 947 303 940 915 2;
  • 60) 0.611 795 947 303 940 915 2 × 2 = 1 + 0.223 591 894 607 881 830 4;
  • 61) 0.223 591 894 607 881 830 4 × 2 = 0 + 0.447 183 789 215 763 660 8;
  • 62) 0.447 183 789 215 763 660 8 × 2 = 0 + 0.894 367 578 431 527 321 6;
  • 63) 0.894 367 578 431 527 321 6 × 2 = 1 + 0.788 735 156 863 054 643 2;
  • 64) 0.788 735 156 863 054 643 2 × 2 = 1 + 0.577 470 313 726 109 286 4;
  • 65) 0.577 470 313 726 109 286 4 × 2 = 1 + 0.154 940 627 452 218 572 8;
  • 66) 0.154 940 627 452 218 572 8 × 2 = 0 + 0.309 881 254 904 437 145 6;
  • 67) 0.309 881 254 904 437 145 6 × 2 = 0 + 0.619 762 509 808 874 291 2;
  • 68) 0.619 762 509 808 874 291 2 × 2 = 1 + 0.239 525 019 617 748 582 4;
  • 69) 0.239 525 019 617 748 582 4 × 2 = 0 + 0.479 050 039 235 497 164 8;
  • 70) 0.479 050 039 235 497 164 8 × 2 = 0 + 0.958 100 078 470 994 329 6;
  • 71) 0.958 100 078 470 994 329 6 × 2 = 1 + 0.916 200 156 941 988 659 2;
  • 72) 0.916 200 156 941 988 659 2 × 2 = 1 + 0.832 400 313 883 977 318 4;
  • 73) 0.832 400 313 883 977 318 4 × 2 = 1 + 0.664 800 627 767 954 636 8;
  • 74) 0.664 800 627 767 954 636 8 × 2 = 1 + 0.329 601 255 535 909 273 6;
  • 75) 0.329 601 255 535 909 273 6 × 2 = 0 + 0.659 202 511 071 818 547 2;
  • 76) 0.659 202 511 071 818 547 2 × 2 = 1 + 0.318 405 022 143 637 094 4;
  • 77) 0.318 405 022 143 637 094 4 × 2 = 0 + 0.636 810 044 287 274 188 8;
  • 78) 0.636 810 044 287 274 188 8 × 2 = 1 + 0.273 620 088 574 548 377 6;
  • 79) 0.273 620 088 574 548 377 6 × 2 = 0 + 0.547 240 177 149 096 755 2;
  • 80) 0.547 240 177 149 096 755 2 × 2 = 1 + 0.094 480 354 298 193 510 4;
  • 81) 0.094 480 354 298 193 510 4 × 2 = 0 + 0.188 960 708 596 387 020 8;
  • 82) 0.188 960 708 596 387 020 8 × 2 = 0 + 0.377 921 417 192 774 041 6;
  • 83) 0.377 921 417 192 774 041 6 × 2 = 0 + 0.755 842 834 385 548 083 2;
  • 84) 0.755 842 834 385 548 083 2 × 2 = 1 + 0.511 685 668 771 096 166 4;
  • 85) 0.511 685 668 771 096 166 4 × 2 = 1 + 0.023 371 337 542 192 332 8;
  • 86) 0.023 371 337 542 192 332 8 × 2 = 0 + 0.046 742 675 084 384 665 6;
  • 87) 0.046 742 675 084 384 665 6 × 2 = 0 + 0.093 485 350 168 769 331 2;
  • 88) 0.093 485 350 168 769 331 2 × 2 = 0 + 0.186 970 700 337 538 662 4;
  • 89) 0.186 970 700 337 538 662 4 × 2 = 0 + 0.373 941 400 675 077 324 8;
  • 90) 0.373 941 400 675 077 324 8 × 2 = 0 + 0.747 882 801 350 154 649 6;
  • 91) 0.747 882 801 350 154 649 6 × 2 = 1 + 0.495 765 602 700 309 299 2;
  • 92) 0.495 765 602 700 309 299 2 × 2 = 0 + 0.991 531 205 400 618 598 4;
  • 93) 0.991 531 205 400 618 598 4 × 2 = 1 + 0.983 062 410 801 237 196 8;
  • 94) 0.983 062 410 801 237 196 8 × 2 = 1 + 0.966 124 821 602 474 393 6;
  • 95) 0.966 124 821 602 474 393 6 × 2 = 1 + 0.932 249 643 204 948 787 2;

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:


0.000 000 000 000 176 540 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0001 0011 1001 0011 1101 0101 0001 1000 0010 111(2)

6. Positive number before normalization:

0.000 000 000 000 176 540 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0001 0011 1001 0011 1101 0101 0001 1000 0010 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:


0.000 000 000 000 176 540 4(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0001 0011 1001 0011 1101 0101 0001 1000 0010 111(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0001 0001 0011 1001 0011 1101 0101 0001 1000 0010 111(2) × 20 =

1.1000 1101 1000 1000 1001 1100 1001 1110 1010 1000 1100 0001 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 1000 1000 1001 1100 1001 1110 1010 1000 1100 0001 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 1000 1000 1001 1100 1001 1110 1010 1000 1100 0001 0111 =

1000 1101 1000 1000 1001 1100 1001 1110 1010 1000 1100 0001 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 1000 1000 1001 1100 1001 1110 1010 1000 1100 0001 0111


Decimal number -0.000 000 000 000 176 540 4 converted to 64 bit double precision IEEE 754 binary floating point representation:

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