-0.000 000 000 000 176 556 73 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 73(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 73(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 73| = 0.000 000 000 000 176 556 73


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 73.

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 73 × 2 = 0 + 0.000 000 000 000 353 113 46;
  • 2) 0.000 000 000 000 353 113 46 × 2 = 0 + 0.000 000 000 000 706 226 92;
  • 3) 0.000 000 000 000 706 226 92 × 2 = 0 + 0.000 000 000 001 412 453 84;
  • 4) 0.000 000 000 001 412 453 84 × 2 = 0 + 0.000 000 000 002 824 907 68;
  • 5) 0.000 000 000 002 824 907 68 × 2 = 0 + 0.000 000 000 005 649 815 36;
  • 6) 0.000 000 000 005 649 815 36 × 2 = 0 + 0.000 000 000 011 299 630 72;
  • 7) 0.000 000 000 011 299 630 72 × 2 = 0 + 0.000 000 000 022 599 261 44;
  • 8) 0.000 000 000 022 599 261 44 × 2 = 0 + 0.000 000 000 045 198 522 88;
  • 9) 0.000 000 000 045 198 522 88 × 2 = 0 + 0.000 000 000 090 397 045 76;
  • 10) 0.000 000 000 090 397 045 76 × 2 = 0 + 0.000 000 000 180 794 091 52;
  • 11) 0.000 000 000 180 794 091 52 × 2 = 0 + 0.000 000 000 361 588 183 04;
  • 12) 0.000 000 000 361 588 183 04 × 2 = 0 + 0.000 000 000 723 176 366 08;
  • 13) 0.000 000 000 723 176 366 08 × 2 = 0 + 0.000 000 001 446 352 732 16;
  • 14) 0.000 000 001 446 352 732 16 × 2 = 0 + 0.000 000 002 892 705 464 32;
  • 15) 0.000 000 002 892 705 464 32 × 2 = 0 + 0.000 000 005 785 410 928 64;
  • 16) 0.000 000 005 785 410 928 64 × 2 = 0 + 0.000 000 011 570 821 857 28;
  • 17) 0.000 000 011 570 821 857 28 × 2 = 0 + 0.000 000 023 141 643 714 56;
  • 18) 0.000 000 023 141 643 714 56 × 2 = 0 + 0.000 000 046 283 287 429 12;
  • 19) 0.000 000 046 283 287 429 12 × 2 = 0 + 0.000 000 092 566 574 858 24;
  • 20) 0.000 000 092 566 574 858 24 × 2 = 0 + 0.000 000 185 133 149 716 48;
  • 21) 0.000 000 185 133 149 716 48 × 2 = 0 + 0.000 000 370 266 299 432 96;
  • 22) 0.000 000 370 266 299 432 96 × 2 = 0 + 0.000 000 740 532 598 865 92;
  • 23) 0.000 000 740 532 598 865 92 × 2 = 0 + 0.000 001 481 065 197 731 84;
  • 24) 0.000 001 481 065 197 731 84 × 2 = 0 + 0.000 002 962 130 395 463 68;
  • 25) 0.000 002 962 130 395 463 68 × 2 = 0 + 0.000 005 924 260 790 927 36;
  • 26) 0.000 005 924 260 790 927 36 × 2 = 0 + 0.000 011 848 521 581 854 72;
  • 27) 0.000 011 848 521 581 854 72 × 2 = 0 + 0.000 023 697 043 163 709 44;
  • 28) 0.000 023 697 043 163 709 44 × 2 = 0 + 0.000 047 394 086 327 418 88;
  • 29) 0.000 047 394 086 327 418 88 × 2 = 0 + 0.000 094 788 172 654 837 76;
  • 30) 0.000 094 788 172 654 837 76 × 2 = 0 + 0.000 189 576 345 309 675 52;
  • 31) 0.000 189 576 345 309 675 52 × 2 = 0 + 0.000 379 152 690 619 351 04;
  • 32) 0.000 379 152 690 619 351 04 × 2 = 0 + 0.000 758 305 381 238 702 08;
  • 33) 0.000 758 305 381 238 702 08 × 2 = 0 + 0.001 516 610 762 477 404 16;
  • 34) 0.001 516 610 762 477 404 16 × 2 = 0 + 0.003 033 221 524 954 808 32;
  • 35) 0.003 033 221 524 954 808 32 × 2 = 0 + 0.006 066 443 049 909 616 64;
  • 36) 0.006 066 443 049 909 616 64 × 2 = 0 + 0.012 132 886 099 819 233 28;
  • 37) 0.012 132 886 099 819 233 28 × 2 = 0 + 0.024 265 772 199 638 466 56;
  • 38) 0.024 265 772 199 638 466 56 × 2 = 0 + 0.048 531 544 399 276 933 12;
  • 39) 0.048 531 544 399 276 933 12 × 2 = 0 + 0.097 063 088 798 553 866 24;
  • 40) 0.097 063 088 798 553 866 24 × 2 = 0 + 0.194 126 177 597 107 732 48;
  • 41) 0.194 126 177 597 107 732 48 × 2 = 0 + 0.388 252 355 194 215 464 96;
  • 42) 0.388 252 355 194 215 464 96 × 2 = 0 + 0.776 504 710 388 430 929 92;
  • 43) 0.776 504 710 388 430 929 92 × 2 = 1 + 0.553 009 420 776 861 859 84;
  • 44) 0.553 009 420 776 861 859 84 × 2 = 1 + 0.106 018 841 553 723 719 68;
  • 45) 0.106 018 841 553 723 719 68 × 2 = 0 + 0.212 037 683 107 447 439 36;
  • 46) 0.212 037 683 107 447 439 36 × 2 = 0 + 0.424 075 366 214 894 878 72;
  • 47) 0.424 075 366 214 894 878 72 × 2 = 0 + 0.848 150 732 429 789 757 44;
  • 48) 0.848 150 732 429 789 757 44 × 2 = 1 + 0.696 301 464 859 579 514 88;
  • 49) 0.696 301 464 859 579 514 88 × 2 = 1 + 0.392 602 929 719 159 029 76;
  • 50) 0.392 602 929 719 159 029 76 × 2 = 0 + 0.785 205 859 438 318 059 52;
  • 51) 0.785 205 859 438 318 059 52 × 2 = 1 + 0.570 411 718 876 636 119 04;
  • 52) 0.570 411 718 876 636 119 04 × 2 = 1 + 0.140 823 437 753 272 238 08;
  • 53) 0.140 823 437 753 272 238 08 × 2 = 0 + 0.281 646 875 506 544 476 16;
  • 54) 0.281 646 875 506 544 476 16 × 2 = 0 + 0.563 293 751 013 088 952 32;
  • 55) 0.563 293 751 013 088 952 32 × 2 = 1 + 0.126 587 502 026 177 904 64;
  • 56) 0.126 587 502 026 177 904 64 × 2 = 0 + 0.253 175 004 052 355 809 28;
  • 57) 0.253 175 004 052 355 809 28 × 2 = 0 + 0.506 350 008 104 711 618 56;
  • 58) 0.506 350 008 104 711 618 56 × 2 = 1 + 0.012 700 016 209 423 237 12;
  • 59) 0.012 700 016 209 423 237 12 × 2 = 0 + 0.025 400 032 418 846 474 24;
  • 60) 0.025 400 032 418 846 474 24 × 2 = 0 + 0.050 800 064 837 692 948 48;
  • 61) 0.050 800 064 837 692 948 48 × 2 = 0 + 0.101 600 129 675 385 896 96;
  • 62) 0.101 600 129 675 385 896 96 × 2 = 0 + 0.203 200 259 350 771 793 92;
  • 63) 0.203 200 259 350 771 793 92 × 2 = 0 + 0.406 400 518 701 543 587 84;
  • 64) 0.406 400 518 701 543 587 84 × 2 = 0 + 0.812 801 037 403 087 175 68;
  • 65) 0.812 801 037 403 087 175 68 × 2 = 1 + 0.625 602 074 806 174 351 36;
  • 66) 0.625 602 074 806 174 351 36 × 2 = 1 + 0.251 204 149 612 348 702 72;
  • 67) 0.251 204 149 612 348 702 72 × 2 = 0 + 0.502 408 299 224 697 405 44;
  • 68) 0.502 408 299 224 697 405 44 × 2 = 1 + 0.004 816 598 449 394 810 88;
  • 69) 0.004 816 598 449 394 810 88 × 2 = 0 + 0.009 633 196 898 789 621 76;
  • 70) 0.009 633 196 898 789 621 76 × 2 = 0 + 0.019 266 393 797 579 243 52;
  • 71) 0.019 266 393 797 579 243 52 × 2 = 0 + 0.038 532 787 595 158 487 04;
  • 72) 0.038 532 787 595 158 487 04 × 2 = 0 + 0.077 065 575 190 316 974 08;
  • 73) 0.077 065 575 190 316 974 08 × 2 = 0 + 0.154 131 150 380 633 948 16;
  • 74) 0.154 131 150 380 633 948 16 × 2 = 0 + 0.308 262 300 761 267 896 32;
  • 75) 0.308 262 300 761 267 896 32 × 2 = 0 + 0.616 524 601 522 535 792 64;
  • 76) 0.616 524 601 522 535 792 64 × 2 = 1 + 0.233 049 203 045 071 585 28;
  • 77) 0.233 049 203 045 071 585 28 × 2 = 0 + 0.466 098 406 090 143 170 56;
  • 78) 0.466 098 406 090 143 170 56 × 2 = 0 + 0.932 196 812 180 286 341 12;
  • 79) 0.932 196 812 180 286 341 12 × 2 = 1 + 0.864 393 624 360 572 682 24;
  • 80) 0.864 393 624 360 572 682 24 × 2 = 1 + 0.728 787 248 721 145 364 48;
  • 81) 0.728 787 248 721 145 364 48 × 2 = 1 + 0.457 574 497 442 290 728 96;
  • 82) 0.457 574 497 442 290 728 96 × 2 = 0 + 0.915 148 994 884 581 457 92;
  • 83) 0.915 148 994 884 581 457 92 × 2 = 1 + 0.830 297 989 769 162 915 84;
  • 84) 0.830 297 989 769 162 915 84 × 2 = 1 + 0.660 595 979 538 325 831 68;
  • 85) 0.660 595 979 538 325 831 68 × 2 = 1 + 0.321 191 959 076 651 663 36;
  • 86) 0.321 191 959 076 651 663 36 × 2 = 0 + 0.642 383 918 153 303 326 72;
  • 87) 0.642 383 918 153 303 326 72 × 2 = 1 + 0.284 767 836 306 606 653 44;
  • 88) 0.284 767 836 306 606 653 44 × 2 = 0 + 0.569 535 672 613 213 306 88;
  • 89) 0.569 535 672 613 213 306 88 × 2 = 1 + 0.139 071 345 226 426 613 76;
  • 90) 0.139 071 345 226 426 613 76 × 2 = 0 + 0.278 142 690 452 853 227 52;
  • 91) 0.278 142 690 452 853 227 52 × 2 = 0 + 0.556 285 380 905 706 455 04;
  • 92) 0.556 285 380 905 706 455 04 × 2 = 1 + 0.112 570 761 811 412 910 08;
  • 93) 0.112 570 761 811 412 910 08 × 2 = 0 + 0.225 141 523 622 825 820 16;
  • 94) 0.225 141 523 622 825 820 16 × 2 = 0 + 0.450 283 047 245 651 640 32;
  • 95) 0.450 283 047 245 651 640 32 × 2 = 0 + 0.900 566 094 491 303 280 64;

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 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1101 0000 0001 0011 1011 1010 1001 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1101 0000 0001 0011 1011 1010 1001 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 73(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1101 0000 0001 0011 1011 1010 1001 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0100 0000 1101 0000 0001 0011 1011 1010 1001 000(2) × 20 =

1.1000 1101 1001 0010 0000 0110 1000 0000 1001 1101 1101 0100 1000(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 0000 0110 1000 0000 1001 1101 1101 0100 1000


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 0000 0110 1000 0000 1001 1101 1101 0100 1000 =

1000 1101 1001 0010 0000 0110 1000 0000 1001 1101 1101 0100 1000


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 0000 0110 1000 0000 1001 1101 1101 0100 1000


Decimal number -0.000 000 000 000 176 556 73 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0010 0000 0110 1000 0000 1001 1101 1101 0100 1000

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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