-0.000 000 000 000 176 556 52 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 52(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 52(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 52| = 0.000 000 000 000 176 556 52


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 52.

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 52 × 2 = 0 + 0.000 000 000 000 353 113 04;
  • 2) 0.000 000 000 000 353 113 04 × 2 = 0 + 0.000 000 000 000 706 226 08;
  • 3) 0.000 000 000 000 706 226 08 × 2 = 0 + 0.000 000 000 001 412 452 16;
  • 4) 0.000 000 000 001 412 452 16 × 2 = 0 + 0.000 000 000 002 824 904 32;
  • 5) 0.000 000 000 002 824 904 32 × 2 = 0 + 0.000 000 000 005 649 808 64;
  • 6) 0.000 000 000 005 649 808 64 × 2 = 0 + 0.000 000 000 011 299 617 28;
  • 7) 0.000 000 000 011 299 617 28 × 2 = 0 + 0.000 000 000 022 599 234 56;
  • 8) 0.000 000 000 022 599 234 56 × 2 = 0 + 0.000 000 000 045 198 469 12;
  • 9) 0.000 000 000 045 198 469 12 × 2 = 0 + 0.000 000 000 090 396 938 24;
  • 10) 0.000 000 000 090 396 938 24 × 2 = 0 + 0.000 000 000 180 793 876 48;
  • 11) 0.000 000 000 180 793 876 48 × 2 = 0 + 0.000 000 000 361 587 752 96;
  • 12) 0.000 000 000 361 587 752 96 × 2 = 0 + 0.000 000 000 723 175 505 92;
  • 13) 0.000 000 000 723 175 505 92 × 2 = 0 + 0.000 000 001 446 351 011 84;
  • 14) 0.000 000 001 446 351 011 84 × 2 = 0 + 0.000 000 002 892 702 023 68;
  • 15) 0.000 000 002 892 702 023 68 × 2 = 0 + 0.000 000 005 785 404 047 36;
  • 16) 0.000 000 005 785 404 047 36 × 2 = 0 + 0.000 000 011 570 808 094 72;
  • 17) 0.000 000 011 570 808 094 72 × 2 = 0 + 0.000 000 023 141 616 189 44;
  • 18) 0.000 000 023 141 616 189 44 × 2 = 0 + 0.000 000 046 283 232 378 88;
  • 19) 0.000 000 046 283 232 378 88 × 2 = 0 + 0.000 000 092 566 464 757 76;
  • 20) 0.000 000 092 566 464 757 76 × 2 = 0 + 0.000 000 185 132 929 515 52;
  • 21) 0.000 000 185 132 929 515 52 × 2 = 0 + 0.000 000 370 265 859 031 04;
  • 22) 0.000 000 370 265 859 031 04 × 2 = 0 + 0.000 000 740 531 718 062 08;
  • 23) 0.000 000 740 531 718 062 08 × 2 = 0 + 0.000 001 481 063 436 124 16;
  • 24) 0.000 001 481 063 436 124 16 × 2 = 0 + 0.000 002 962 126 872 248 32;
  • 25) 0.000 002 962 126 872 248 32 × 2 = 0 + 0.000 005 924 253 744 496 64;
  • 26) 0.000 005 924 253 744 496 64 × 2 = 0 + 0.000 011 848 507 488 993 28;
  • 27) 0.000 011 848 507 488 993 28 × 2 = 0 + 0.000 023 697 014 977 986 56;
  • 28) 0.000 023 697 014 977 986 56 × 2 = 0 + 0.000 047 394 029 955 973 12;
  • 29) 0.000 047 394 029 955 973 12 × 2 = 0 + 0.000 094 788 059 911 946 24;
  • 30) 0.000 094 788 059 911 946 24 × 2 = 0 + 0.000 189 576 119 823 892 48;
  • 31) 0.000 189 576 119 823 892 48 × 2 = 0 + 0.000 379 152 239 647 784 96;
  • 32) 0.000 379 152 239 647 784 96 × 2 = 0 + 0.000 758 304 479 295 569 92;
  • 33) 0.000 758 304 479 295 569 92 × 2 = 0 + 0.001 516 608 958 591 139 84;
  • 34) 0.001 516 608 958 591 139 84 × 2 = 0 + 0.003 033 217 917 182 279 68;
  • 35) 0.003 033 217 917 182 279 68 × 2 = 0 + 0.006 066 435 834 364 559 36;
  • 36) 0.006 066 435 834 364 559 36 × 2 = 0 + 0.012 132 871 668 729 118 72;
  • 37) 0.012 132 871 668 729 118 72 × 2 = 0 + 0.024 265 743 337 458 237 44;
  • 38) 0.024 265 743 337 458 237 44 × 2 = 0 + 0.048 531 486 674 916 474 88;
  • 39) 0.048 531 486 674 916 474 88 × 2 = 0 + 0.097 062 973 349 832 949 76;
  • 40) 0.097 062 973 349 832 949 76 × 2 = 0 + 0.194 125 946 699 665 899 52;
  • 41) 0.194 125 946 699 665 899 52 × 2 = 0 + 0.388 251 893 399 331 799 04;
  • 42) 0.388 251 893 399 331 799 04 × 2 = 0 + 0.776 503 786 798 663 598 08;
  • 43) 0.776 503 786 798 663 598 08 × 2 = 1 + 0.553 007 573 597 327 196 16;
  • 44) 0.553 007 573 597 327 196 16 × 2 = 1 + 0.106 015 147 194 654 392 32;
  • 45) 0.106 015 147 194 654 392 32 × 2 = 0 + 0.212 030 294 389 308 784 64;
  • 46) 0.212 030 294 389 308 784 64 × 2 = 0 + 0.424 060 588 778 617 569 28;
  • 47) 0.424 060 588 778 617 569 28 × 2 = 0 + 0.848 121 177 557 235 138 56;
  • 48) 0.848 121 177 557 235 138 56 × 2 = 1 + 0.696 242 355 114 470 277 12;
  • 49) 0.696 242 355 114 470 277 12 × 2 = 1 + 0.392 484 710 228 940 554 24;
  • 50) 0.392 484 710 228 940 554 24 × 2 = 0 + 0.784 969 420 457 881 108 48;
  • 51) 0.784 969 420 457 881 108 48 × 2 = 1 + 0.569 938 840 915 762 216 96;
  • 52) 0.569 938 840 915 762 216 96 × 2 = 1 + 0.139 877 681 831 524 433 92;
  • 53) 0.139 877 681 831 524 433 92 × 2 = 0 + 0.279 755 363 663 048 867 84;
  • 54) 0.279 755 363 663 048 867 84 × 2 = 0 + 0.559 510 727 326 097 735 68;
  • 55) 0.559 510 727 326 097 735 68 × 2 = 1 + 0.119 021 454 652 195 471 36;
  • 56) 0.119 021 454 652 195 471 36 × 2 = 0 + 0.238 042 909 304 390 942 72;
  • 57) 0.238 042 909 304 390 942 72 × 2 = 0 + 0.476 085 818 608 781 885 44;
  • 58) 0.476 085 818 608 781 885 44 × 2 = 0 + 0.952 171 637 217 563 770 88;
  • 59) 0.952 171 637 217 563 770 88 × 2 = 1 + 0.904 343 274 435 127 541 76;
  • 60) 0.904 343 274 435 127 541 76 × 2 = 1 + 0.808 686 548 870 255 083 52;
  • 61) 0.808 686 548 870 255 083 52 × 2 = 1 + 0.617 373 097 740 510 167 04;
  • 62) 0.617 373 097 740 510 167 04 × 2 = 1 + 0.234 746 195 481 020 334 08;
  • 63) 0.234 746 195 481 020 334 08 × 2 = 0 + 0.469 492 390 962 040 668 16;
  • 64) 0.469 492 390 962 040 668 16 × 2 = 0 + 0.938 984 781 924 081 336 32;
  • 65) 0.938 984 781 924 081 336 32 × 2 = 1 + 0.877 969 563 848 162 672 64;
  • 66) 0.877 969 563 848 162 672 64 × 2 = 1 + 0.755 939 127 696 325 345 28;
  • 67) 0.755 939 127 696 325 345 28 × 2 = 1 + 0.511 878 255 392 650 690 56;
  • 68) 0.511 878 255 392 650 690 56 × 2 = 1 + 0.023 756 510 785 301 381 12;
  • 69) 0.023 756 510 785 301 381 12 × 2 = 0 + 0.047 513 021 570 602 762 24;
  • 70) 0.047 513 021 570 602 762 24 × 2 = 0 + 0.095 026 043 141 205 524 48;
  • 71) 0.095 026 043 141 205 524 48 × 2 = 0 + 0.190 052 086 282 411 048 96;
  • 72) 0.190 052 086 282 411 048 96 × 2 = 0 + 0.380 104 172 564 822 097 92;
  • 73) 0.380 104 172 564 822 097 92 × 2 = 0 + 0.760 208 345 129 644 195 84;
  • 74) 0.760 208 345 129 644 195 84 × 2 = 1 + 0.520 416 690 259 288 391 68;
  • 75) 0.520 416 690 259 288 391 68 × 2 = 1 + 0.040 833 380 518 576 783 36;
  • 76) 0.040 833 380 518 576 783 36 × 2 = 0 + 0.081 666 761 037 153 566 72;
  • 77) 0.081 666 761 037 153 566 72 × 2 = 0 + 0.163 333 522 074 307 133 44;
  • 78) 0.163 333 522 074 307 133 44 × 2 = 0 + 0.326 667 044 148 614 266 88;
  • 79) 0.326 667 044 148 614 266 88 × 2 = 0 + 0.653 334 088 297 228 533 76;
  • 80) 0.653 334 088 297 228 533 76 × 2 = 1 + 0.306 668 176 594 457 067 52;
  • 81) 0.306 668 176 594 457 067 52 × 2 = 0 + 0.613 336 353 188 914 135 04;
  • 82) 0.613 336 353 188 914 135 04 × 2 = 1 + 0.226 672 706 377 828 270 08;
  • 83) 0.226 672 706 377 828 270 08 × 2 = 0 + 0.453 345 412 755 656 540 16;
  • 84) 0.453 345 412 755 656 540 16 × 2 = 0 + 0.906 690 825 511 313 080 32;
  • 85) 0.906 690 825 511 313 080 32 × 2 = 1 + 0.813 381 651 022 626 160 64;
  • 86) 0.813 381 651 022 626 160 64 × 2 = 1 + 0.626 763 302 045 252 321 28;
  • 87) 0.626 763 302 045 252 321 28 × 2 = 1 + 0.253 526 604 090 504 642 56;
  • 88) 0.253 526 604 090 504 642 56 × 2 = 0 + 0.507 053 208 181 009 285 12;
  • 89) 0.507 053 208 181 009 285 12 × 2 = 1 + 0.014 106 416 362 018 570 24;
  • 90) 0.014 106 416 362 018 570 24 × 2 = 0 + 0.028 212 832 724 037 140 48;
  • 91) 0.028 212 832 724 037 140 48 × 2 = 0 + 0.056 425 665 448 074 280 96;
  • 92) 0.056 425 665 448 074 280 96 × 2 = 0 + 0.112 851 330 896 148 561 92;
  • 93) 0.112 851 330 896 148 561 92 × 2 = 0 + 0.225 702 661 792 297 123 84;
  • 94) 0.225 702 661 792 297 123 84 × 2 = 0 + 0.451 405 323 584 594 247 68;
  • 95) 0.451 405 323 584 594 247 68 × 2 = 0 + 0.902 810 647 169 188 495 36;

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 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1100 1111 0000 0110 0001 0100 1110 1000 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1100 1111 0000 0110 0001 0100 1110 1000 000(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 52(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1100 1111 0000 0110 0001 0100 1110 1000 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1100 1111 0000 0110 0001 0100 1110 1000 000(2) × 20 =

1.1000 1101 1001 0001 1110 0111 1000 0011 0000 1010 0111 0100 0000(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 1110 0111 1000 0011 0000 1010 0111 0100 0000


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 1110 0111 1000 0011 0000 1010 0111 0100 0000 =

1000 1101 1001 0001 1110 0111 1000 0011 0000 1010 0111 0100 0000


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 1110 0111 1000 0011 0000 1010 0111 0100 0000


Decimal number -0.000 000 000 000 176 556 52 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1110 0111 1000 0011 0000 1010 0111 0100 0000

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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