-0.381 966 011 255 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 255 11(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
-0.381 966 011 255 11(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:

|-0.381 966 011 255 11| = 0.381 966 011 255 11


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: 0.381 966 011 255 11.

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) 0.381 966 011 255 11 × 2 = 0 + 0.763 932 022 510 22;
  • 2) 0.763 932 022 510 22 × 2 = 1 + 0.527 864 045 020 44;
  • 3) 0.527 864 045 020 44 × 2 = 1 + 0.055 728 090 040 88;
  • 4) 0.055 728 090 040 88 × 2 = 0 + 0.111 456 180 081 76;
  • 5) 0.111 456 180 081 76 × 2 = 0 + 0.222 912 360 163 52;
  • 6) 0.222 912 360 163 52 × 2 = 0 + 0.445 824 720 327 04;
  • 7) 0.445 824 720 327 04 × 2 = 0 + 0.891 649 440 654 08;
  • 8) 0.891 649 440 654 08 × 2 = 1 + 0.783 298 881 308 16;
  • 9) 0.783 298 881 308 16 × 2 = 1 + 0.566 597 762 616 32;
  • 10) 0.566 597 762 616 32 × 2 = 1 + 0.133 195 525 232 64;
  • 11) 0.133 195 525 232 64 × 2 = 0 + 0.266 391 050 465 28;
  • 12) 0.266 391 050 465 28 × 2 = 0 + 0.532 782 100 930 56;
  • 13) 0.532 782 100 930 56 × 2 = 1 + 0.065 564 201 861 12;
  • 14) 0.065 564 201 861 12 × 2 = 0 + 0.131 128 403 722 24;
  • 15) 0.131 128 403 722 24 × 2 = 0 + 0.262 256 807 444 48;
  • 16) 0.262 256 807 444 48 × 2 = 0 + 0.524 513 614 888 96;
  • 17) 0.524 513 614 888 96 × 2 = 1 + 0.049 027 229 777 92;
  • 18) 0.049 027 229 777 92 × 2 = 0 + 0.098 054 459 555 84;
  • 19) 0.098 054 459 555 84 × 2 = 0 + 0.196 108 919 111 68;
  • 20) 0.196 108 919 111 68 × 2 = 0 + 0.392 217 838 223 36;
  • 21) 0.392 217 838 223 36 × 2 = 0 + 0.784 435 676 446 72;
  • 22) 0.784 435 676 446 72 × 2 = 1 + 0.568 871 352 893 44;
  • 23) 0.568 871 352 893 44 × 2 = 1 + 0.137 742 705 786 88;
  • 24) 0.137 742 705 786 88 × 2 = 0 + 0.275 485 411 573 76;
  • 25) 0.275 485 411 573 76 × 2 = 0 + 0.550 970 823 147 52;
  • 26) 0.550 970 823 147 52 × 2 = 1 + 0.101 941 646 295 04;
  • 27) 0.101 941 646 295 04 × 2 = 0 + 0.203 883 292 590 08;
  • 28) 0.203 883 292 590 08 × 2 = 0 + 0.407 766 585 180 16;
  • 29) 0.407 766 585 180 16 × 2 = 0 + 0.815 533 170 360 32;
  • 30) 0.815 533 170 360 32 × 2 = 1 + 0.631 066 340 720 64;
  • 31) 0.631 066 340 720 64 × 2 = 1 + 0.262 132 681 441 28;
  • 32) 0.262 132 681 441 28 × 2 = 0 + 0.524 265 362 882 56;
  • 33) 0.524 265 362 882 56 × 2 = 1 + 0.048 530 725 765 12;
  • 34) 0.048 530 725 765 12 × 2 = 0 + 0.097 061 451 530 24;
  • 35) 0.097 061 451 530 24 × 2 = 0 + 0.194 122 903 060 48;
  • 36) 0.194 122 903 060 48 × 2 = 0 + 0.388 245 806 120 96;
  • 37) 0.388 245 806 120 96 × 2 = 0 + 0.776 491 612 241 92;
  • 38) 0.776 491 612 241 92 × 2 = 1 + 0.552 983 224 483 84;
  • 39) 0.552 983 224 483 84 × 2 = 1 + 0.105 966 448 967 68;
  • 40) 0.105 966 448 967 68 × 2 = 0 + 0.211 932 897 935 36;
  • 41) 0.211 932 897 935 36 × 2 = 0 + 0.423 865 795 870 72;
  • 42) 0.423 865 795 870 72 × 2 = 0 + 0.847 731 591 741 44;
  • 43) 0.847 731 591 741 44 × 2 = 1 + 0.695 463 183 482 88;
  • 44) 0.695 463 183 482 88 × 2 = 1 + 0.390 926 366 965 76;
  • 45) 0.390 926 366 965 76 × 2 = 0 + 0.781 852 733 931 52;
  • 46) 0.781 852 733 931 52 × 2 = 1 + 0.563 705 467 863 04;
  • 47) 0.563 705 467 863 04 × 2 = 1 + 0.127 410 935 726 08;
  • 48) 0.127 410 935 726 08 × 2 = 0 + 0.254 821 871 452 16;
  • 49) 0.254 821 871 452 16 × 2 = 0 + 0.509 643 742 904 32;
  • 50) 0.509 643 742 904 32 × 2 = 1 + 0.019 287 485 808 64;
  • 51) 0.019 287 485 808 64 × 2 = 0 + 0.038 574 971 617 28;
  • 52) 0.038 574 971 617 28 × 2 = 0 + 0.077 149 943 234 56;
  • 53) 0.077 149 943 234 56 × 2 = 0 + 0.154 299 886 469 12;
  • 54) 0.154 299 886 469 12 × 2 = 0 + 0.308 599 772 938 24;

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:


0.381 966 011 255 11(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0011 0110 0100 00(2)

6. Positive number before normalization:

0.381 966 011 255 11(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0011 0110 0100 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 2 positions to the right, so that only one non zero digit remains to the left of it:


0.381 966 011 255 11(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0011 0110 0100 00(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0011 0110 0100 00(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 1001 0000(2) × 2-2


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): -2

Mantissa (not normalized):
1.1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 1001 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-2 + 2(11-1) - 1 =

(-2 + 1 023)(10) =

1 021(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;
  • 1 021 ÷ 2 = 510 + 1;
  • 510 ÷ 2 = 255 + 0;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

1021(10) =

011 1111 1101(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 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 1001 0000 =

1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 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 1111 1101

Mantissa (52 bits) =
1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 1001 0000


Decimal number -0.381 966 011 255 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 1101 - 1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 1101 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