-0.000 000 000 000 176 556 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 176 556 38(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 38(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 38| = 0.000 000 000 000 176 556 38


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 38.

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 38 × 2 = 0 + 0.000 000 000 000 353 112 76;
  • 2) 0.000 000 000 000 353 112 76 × 2 = 0 + 0.000 000 000 000 706 225 52;
  • 3) 0.000 000 000 000 706 225 52 × 2 = 0 + 0.000 000 000 001 412 451 04;
  • 4) 0.000 000 000 001 412 451 04 × 2 = 0 + 0.000 000 000 002 824 902 08;
  • 5) 0.000 000 000 002 824 902 08 × 2 = 0 + 0.000 000 000 005 649 804 16;
  • 6) 0.000 000 000 005 649 804 16 × 2 = 0 + 0.000 000 000 011 299 608 32;
  • 7) 0.000 000 000 011 299 608 32 × 2 = 0 + 0.000 000 000 022 599 216 64;
  • 8) 0.000 000 000 022 599 216 64 × 2 = 0 + 0.000 000 000 045 198 433 28;
  • 9) 0.000 000 000 045 198 433 28 × 2 = 0 + 0.000 000 000 090 396 866 56;
  • 10) 0.000 000 000 090 396 866 56 × 2 = 0 + 0.000 000 000 180 793 733 12;
  • 11) 0.000 000 000 180 793 733 12 × 2 = 0 + 0.000 000 000 361 587 466 24;
  • 12) 0.000 000 000 361 587 466 24 × 2 = 0 + 0.000 000 000 723 174 932 48;
  • 13) 0.000 000 000 723 174 932 48 × 2 = 0 + 0.000 000 001 446 349 864 96;
  • 14) 0.000 000 001 446 349 864 96 × 2 = 0 + 0.000 000 002 892 699 729 92;
  • 15) 0.000 000 002 892 699 729 92 × 2 = 0 + 0.000 000 005 785 399 459 84;
  • 16) 0.000 000 005 785 399 459 84 × 2 = 0 + 0.000 000 011 570 798 919 68;
  • 17) 0.000 000 011 570 798 919 68 × 2 = 0 + 0.000 000 023 141 597 839 36;
  • 18) 0.000 000 023 141 597 839 36 × 2 = 0 + 0.000 000 046 283 195 678 72;
  • 19) 0.000 000 046 283 195 678 72 × 2 = 0 + 0.000 000 092 566 391 357 44;
  • 20) 0.000 000 092 566 391 357 44 × 2 = 0 + 0.000 000 185 132 782 714 88;
  • 21) 0.000 000 185 132 782 714 88 × 2 = 0 + 0.000 000 370 265 565 429 76;
  • 22) 0.000 000 370 265 565 429 76 × 2 = 0 + 0.000 000 740 531 130 859 52;
  • 23) 0.000 000 740 531 130 859 52 × 2 = 0 + 0.000 001 481 062 261 719 04;
  • 24) 0.000 001 481 062 261 719 04 × 2 = 0 + 0.000 002 962 124 523 438 08;
  • 25) 0.000 002 962 124 523 438 08 × 2 = 0 + 0.000 005 924 249 046 876 16;
  • 26) 0.000 005 924 249 046 876 16 × 2 = 0 + 0.000 011 848 498 093 752 32;
  • 27) 0.000 011 848 498 093 752 32 × 2 = 0 + 0.000 023 696 996 187 504 64;
  • 28) 0.000 023 696 996 187 504 64 × 2 = 0 + 0.000 047 393 992 375 009 28;
  • 29) 0.000 047 393 992 375 009 28 × 2 = 0 + 0.000 094 787 984 750 018 56;
  • 30) 0.000 094 787 984 750 018 56 × 2 = 0 + 0.000 189 575 969 500 037 12;
  • 31) 0.000 189 575 969 500 037 12 × 2 = 0 + 0.000 379 151 939 000 074 24;
  • 32) 0.000 379 151 939 000 074 24 × 2 = 0 + 0.000 758 303 878 000 148 48;
  • 33) 0.000 758 303 878 000 148 48 × 2 = 0 + 0.001 516 607 756 000 296 96;
  • 34) 0.001 516 607 756 000 296 96 × 2 = 0 + 0.003 033 215 512 000 593 92;
  • 35) 0.003 033 215 512 000 593 92 × 2 = 0 + 0.006 066 431 024 001 187 84;
  • 36) 0.006 066 431 024 001 187 84 × 2 = 0 + 0.012 132 862 048 002 375 68;
  • 37) 0.012 132 862 048 002 375 68 × 2 = 0 + 0.024 265 724 096 004 751 36;
  • 38) 0.024 265 724 096 004 751 36 × 2 = 0 + 0.048 531 448 192 009 502 72;
  • 39) 0.048 531 448 192 009 502 72 × 2 = 0 + 0.097 062 896 384 019 005 44;
  • 40) 0.097 062 896 384 019 005 44 × 2 = 0 + 0.194 125 792 768 038 010 88;
  • 41) 0.194 125 792 768 038 010 88 × 2 = 0 + 0.388 251 585 536 076 021 76;
  • 42) 0.388 251 585 536 076 021 76 × 2 = 0 + 0.776 503 171 072 152 043 52;
  • 43) 0.776 503 171 072 152 043 52 × 2 = 1 + 0.553 006 342 144 304 087 04;
  • 44) 0.553 006 342 144 304 087 04 × 2 = 1 + 0.106 012 684 288 608 174 08;
  • 45) 0.106 012 684 288 608 174 08 × 2 = 0 + 0.212 025 368 577 216 348 16;
  • 46) 0.212 025 368 577 216 348 16 × 2 = 0 + 0.424 050 737 154 432 696 32;
  • 47) 0.424 050 737 154 432 696 32 × 2 = 0 + 0.848 101 474 308 865 392 64;
  • 48) 0.848 101 474 308 865 392 64 × 2 = 1 + 0.696 202 948 617 730 785 28;
  • 49) 0.696 202 948 617 730 785 28 × 2 = 1 + 0.392 405 897 235 461 570 56;
  • 50) 0.392 405 897 235 461 570 56 × 2 = 0 + 0.784 811 794 470 923 141 12;
  • 51) 0.784 811 794 470 923 141 12 × 2 = 1 + 0.569 623 588 941 846 282 24;
  • 52) 0.569 623 588 941 846 282 24 × 2 = 1 + 0.139 247 177 883 692 564 48;
  • 53) 0.139 247 177 883 692 564 48 × 2 = 0 + 0.278 494 355 767 385 128 96;
  • 54) 0.278 494 355 767 385 128 96 × 2 = 0 + 0.556 988 711 534 770 257 92;
  • 55) 0.556 988 711 534 770 257 92 × 2 = 1 + 0.113 977 423 069 540 515 84;
  • 56) 0.113 977 423 069 540 515 84 × 2 = 0 + 0.227 954 846 139 081 031 68;
  • 57) 0.227 954 846 139 081 031 68 × 2 = 0 + 0.455 909 692 278 162 063 36;
  • 58) 0.455 909 692 278 162 063 36 × 2 = 0 + 0.911 819 384 556 324 126 72;
  • 59) 0.911 819 384 556 324 126 72 × 2 = 1 + 0.823 638 769 112 648 253 44;
  • 60) 0.823 638 769 112 648 253 44 × 2 = 1 + 0.647 277 538 225 296 506 88;
  • 61) 0.647 277 538 225 296 506 88 × 2 = 1 + 0.294 555 076 450 593 013 76;
  • 62) 0.294 555 076 450 593 013 76 × 2 = 0 + 0.589 110 152 901 186 027 52;
  • 63) 0.589 110 152 901 186 027 52 × 2 = 1 + 0.178 220 305 802 372 055 04;
  • 64) 0.178 220 305 802 372 055 04 × 2 = 0 + 0.356 440 611 604 744 110 08;
  • 65) 0.356 440 611 604 744 110 08 × 2 = 0 + 0.712 881 223 209 488 220 16;
  • 66) 0.712 881 223 209 488 220 16 × 2 = 1 + 0.425 762 446 418 976 440 32;
  • 67) 0.425 762 446 418 976 440 32 × 2 = 0 + 0.851 524 892 837 952 880 64;
  • 68) 0.851 524 892 837 952 880 64 × 2 = 1 + 0.703 049 785 675 905 761 28;
  • 69) 0.703 049 785 675 905 761 28 × 2 = 1 + 0.406 099 571 351 811 522 56;
  • 70) 0.406 099 571 351 811 522 56 × 2 = 0 + 0.812 199 142 703 623 045 12;
  • 71) 0.812 199 142 703 623 045 12 × 2 = 1 + 0.624 398 285 407 246 090 24;
  • 72) 0.624 398 285 407 246 090 24 × 2 = 1 + 0.248 796 570 814 492 180 48;
  • 73) 0.248 796 570 814 492 180 48 × 2 = 0 + 0.497 593 141 628 984 360 96;
  • 74) 0.497 593 141 628 984 360 96 × 2 = 0 + 0.995 186 283 257 968 721 92;
  • 75) 0.995 186 283 257 968 721 92 × 2 = 1 + 0.990 372 566 515 937 443 84;
  • 76) 0.990 372 566 515 937 443 84 × 2 = 1 + 0.980 745 133 031 874 887 68;
  • 77) 0.980 745 133 031 874 887 68 × 2 = 1 + 0.961 490 266 063 749 775 36;
  • 78) 0.961 490 266 063 749 775 36 × 2 = 1 + 0.922 980 532 127 499 550 72;
  • 79) 0.922 980 532 127 499 550 72 × 2 = 1 + 0.845 961 064 254 999 101 44;
  • 80) 0.845 961 064 254 999 101 44 × 2 = 1 + 0.691 922 128 509 998 202 88;
  • 81) 0.691 922 128 509 998 202 88 × 2 = 1 + 0.383 844 257 019 996 405 76;
  • 82) 0.383 844 257 019 996 405 76 × 2 = 0 + 0.767 688 514 039 992 811 52;
  • 83) 0.767 688 514 039 992 811 52 × 2 = 1 + 0.535 377 028 079 985 623 04;
  • 84) 0.535 377 028 079 985 623 04 × 2 = 1 + 0.070 754 056 159 971 246 08;
  • 85) 0.070 754 056 159 971 246 08 × 2 = 0 + 0.141 508 112 319 942 492 16;
  • 86) 0.141 508 112 319 942 492 16 × 2 = 0 + 0.283 016 224 639 884 984 32;
  • 87) 0.283 016 224 639 884 984 32 × 2 = 0 + 0.566 032 449 279 769 968 64;
  • 88) 0.566 032 449 279 769 968 64 × 2 = 1 + 0.132 064 898 559 539 937 28;
  • 89) 0.132 064 898 559 539 937 28 × 2 = 0 + 0.264 129 797 119 079 874 56;
  • 90) 0.264 129 797 119 079 874 56 × 2 = 0 + 0.528 259 594 238 159 749 12;
  • 91) 0.528 259 594 238 159 749 12 × 2 = 1 + 0.056 519 188 476 319 498 24;
  • 92) 0.056 519 188 476 319 498 24 × 2 = 0 + 0.113 038 376 952 638 996 48;
  • 93) 0.113 038 376 952 638 996 48 × 2 = 0 + 0.226 076 753 905 277 992 96;
  • 94) 0.226 076 753 905 277 992 96 × 2 = 0 + 0.452 153 507 810 555 985 92;
  • 95) 0.452 153 507 810 555 985 92 × 2 = 0 + 0.904 307 015 621 111 971 84;

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 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1010 0101 1011 0011 1111 1011 0001 0010 000(2)

6. Positive number before normalization:

0.000 000 000 000 176 556 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1010 0101 1011 0011 1111 1011 0001 0010 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 38(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1010 0101 1011 0011 1111 1011 0001 0010 000(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0001 1011 0010 0011 1010 0101 1011 0011 1111 1011 0001 0010 000(2) × 20 =

1.1000 1101 1001 0001 1101 0010 1101 1001 1111 1101 1000 1001 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 1101 0010 1101 1001 1111 1101 1000 1001 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 1101 0010 1101 1001 1111 1101 1000 1001 0000 =

1000 1101 1001 0001 1101 0010 1101 1001 1111 1101 1000 1001 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 1101 0010 1101 1001 1111 1101 1000 1001 0000


Decimal number -0.000 000 000 000 176 556 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1101 0100 - 1000 1101 1001 0001 1101 0010 1101 1001 1111 1101 1000 1001 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