-0.381 966 011 255 02 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.381 966 011 255 02(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 02(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 02| = 0.381 966 011 255 02


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 02.

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 02 × 2 = 0 + 0.763 932 022 510 04;
  • 2) 0.763 932 022 510 04 × 2 = 1 + 0.527 864 045 020 08;
  • 3) 0.527 864 045 020 08 × 2 = 1 + 0.055 728 090 040 16;
  • 4) 0.055 728 090 040 16 × 2 = 0 + 0.111 456 180 080 32;
  • 5) 0.111 456 180 080 32 × 2 = 0 + 0.222 912 360 160 64;
  • 6) 0.222 912 360 160 64 × 2 = 0 + 0.445 824 720 321 28;
  • 7) 0.445 824 720 321 28 × 2 = 0 + 0.891 649 440 642 56;
  • 8) 0.891 649 440 642 56 × 2 = 1 + 0.783 298 881 285 12;
  • 9) 0.783 298 881 285 12 × 2 = 1 + 0.566 597 762 570 24;
  • 10) 0.566 597 762 570 24 × 2 = 1 + 0.133 195 525 140 48;
  • 11) 0.133 195 525 140 48 × 2 = 0 + 0.266 391 050 280 96;
  • 12) 0.266 391 050 280 96 × 2 = 0 + 0.532 782 100 561 92;
  • 13) 0.532 782 100 561 92 × 2 = 1 + 0.065 564 201 123 84;
  • 14) 0.065 564 201 123 84 × 2 = 0 + 0.131 128 402 247 68;
  • 15) 0.131 128 402 247 68 × 2 = 0 + 0.262 256 804 495 36;
  • 16) 0.262 256 804 495 36 × 2 = 0 + 0.524 513 608 990 72;
  • 17) 0.524 513 608 990 72 × 2 = 1 + 0.049 027 217 981 44;
  • 18) 0.049 027 217 981 44 × 2 = 0 + 0.098 054 435 962 88;
  • 19) 0.098 054 435 962 88 × 2 = 0 + 0.196 108 871 925 76;
  • 20) 0.196 108 871 925 76 × 2 = 0 + 0.392 217 743 851 52;
  • 21) 0.392 217 743 851 52 × 2 = 0 + 0.784 435 487 703 04;
  • 22) 0.784 435 487 703 04 × 2 = 1 + 0.568 870 975 406 08;
  • 23) 0.568 870 975 406 08 × 2 = 1 + 0.137 741 950 812 16;
  • 24) 0.137 741 950 812 16 × 2 = 0 + 0.275 483 901 624 32;
  • 25) 0.275 483 901 624 32 × 2 = 0 + 0.550 967 803 248 64;
  • 26) 0.550 967 803 248 64 × 2 = 1 + 0.101 935 606 497 28;
  • 27) 0.101 935 606 497 28 × 2 = 0 + 0.203 871 212 994 56;
  • 28) 0.203 871 212 994 56 × 2 = 0 + 0.407 742 425 989 12;
  • 29) 0.407 742 425 989 12 × 2 = 0 + 0.815 484 851 978 24;
  • 30) 0.815 484 851 978 24 × 2 = 1 + 0.630 969 703 956 48;
  • 31) 0.630 969 703 956 48 × 2 = 1 + 0.261 939 407 912 96;
  • 32) 0.261 939 407 912 96 × 2 = 0 + 0.523 878 815 825 92;
  • 33) 0.523 878 815 825 92 × 2 = 1 + 0.047 757 631 651 84;
  • 34) 0.047 757 631 651 84 × 2 = 0 + 0.095 515 263 303 68;
  • 35) 0.095 515 263 303 68 × 2 = 0 + 0.191 030 526 607 36;
  • 36) 0.191 030 526 607 36 × 2 = 0 + 0.382 061 053 214 72;
  • 37) 0.382 061 053 214 72 × 2 = 0 + 0.764 122 106 429 44;
  • 38) 0.764 122 106 429 44 × 2 = 1 + 0.528 244 212 858 88;
  • 39) 0.528 244 212 858 88 × 2 = 1 + 0.056 488 425 717 76;
  • 40) 0.056 488 425 717 76 × 2 = 0 + 0.112 976 851 435 52;
  • 41) 0.112 976 851 435 52 × 2 = 0 + 0.225 953 702 871 04;
  • 42) 0.225 953 702 871 04 × 2 = 0 + 0.451 907 405 742 08;
  • 43) 0.451 907 405 742 08 × 2 = 0 + 0.903 814 811 484 16;
  • 44) 0.903 814 811 484 16 × 2 = 1 + 0.807 629 622 968 32;
  • 45) 0.807 629 622 968 32 × 2 = 1 + 0.615 259 245 936 64;
  • 46) 0.615 259 245 936 64 × 2 = 1 + 0.230 518 491 873 28;
  • 47) 0.230 518 491 873 28 × 2 = 0 + 0.461 036 983 746 56;
  • 48) 0.461 036 983 746 56 × 2 = 0 + 0.922 073 967 493 12;
  • 49) 0.922 073 967 493 12 × 2 = 1 + 0.844 147 934 986 24;
  • 50) 0.844 147 934 986 24 × 2 = 1 + 0.688 295 869 972 48;
  • 51) 0.688 295 869 972 48 × 2 = 1 + 0.376 591 739 944 96;
  • 52) 0.376 591 739 944 96 × 2 = 0 + 0.753 183 479 889 92;
  • 53) 0.753 183 479 889 92 × 2 = 1 + 0.506 366 959 779 84;
  • 54) 0.506 366 959 779 84 × 2 = 1 + 0.012 733 919 559 68;

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 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0001 1100 1110 11(2)

6. Positive number before normalization:

0.381 966 011 255 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0001 1100 1110 11(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 02(10) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0001 1100 1110 11(2) =

0.0110 0001 1100 1000 1000 0110 0100 0110 1000 0110 0001 1100 1110 11(2) × 20 =

1.1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 0111 0011 1011(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 0111 0011 1011


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 0111 0011 1011 =

1000 0111 0010 0010 0001 1001 0001 1010 0001 1000 0111 0011 1011


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 0111 0011 1011


Decimal number -0.381 966 011 255 02 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 0111 0011 1011

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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