-0.000 283 11 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 283 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
-0.000 283 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:

|-0.000 283 11| = 0.000 283 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: 0.000 283 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) 0.000 283 11 × 2 = 0 + 0.000 566 22;
  • 2) 0.000 566 22 × 2 = 0 + 0.001 132 44;
  • 3) 0.001 132 44 × 2 = 0 + 0.002 264 88;
  • 4) 0.002 264 88 × 2 = 0 + 0.004 529 76;
  • 5) 0.004 529 76 × 2 = 0 + 0.009 059 52;
  • 6) 0.009 059 52 × 2 = 0 + 0.018 119 04;
  • 7) 0.018 119 04 × 2 = 0 + 0.036 238 08;
  • 8) 0.036 238 08 × 2 = 0 + 0.072 476 16;
  • 9) 0.072 476 16 × 2 = 0 + 0.144 952 32;
  • 10) 0.144 952 32 × 2 = 0 + 0.289 904 64;
  • 11) 0.289 904 64 × 2 = 0 + 0.579 809 28;
  • 12) 0.579 809 28 × 2 = 1 + 0.159 618 56;
  • 13) 0.159 618 56 × 2 = 0 + 0.319 237 12;
  • 14) 0.319 237 12 × 2 = 0 + 0.638 474 24;
  • 15) 0.638 474 24 × 2 = 1 + 0.276 948 48;
  • 16) 0.276 948 48 × 2 = 0 + 0.553 896 96;
  • 17) 0.553 896 96 × 2 = 1 + 0.107 793 92;
  • 18) 0.107 793 92 × 2 = 0 + 0.215 587 84;
  • 19) 0.215 587 84 × 2 = 0 + 0.431 175 68;
  • 20) 0.431 175 68 × 2 = 0 + 0.862 351 36;
  • 21) 0.862 351 36 × 2 = 1 + 0.724 702 72;
  • 22) 0.724 702 72 × 2 = 1 + 0.449 405 44;
  • 23) 0.449 405 44 × 2 = 0 + 0.898 810 88;
  • 24) 0.898 810 88 × 2 = 1 + 0.797 621 76;
  • 25) 0.797 621 76 × 2 = 1 + 0.595 243 52;
  • 26) 0.595 243 52 × 2 = 1 + 0.190 487 04;
  • 27) 0.190 487 04 × 2 = 0 + 0.380 974 08;
  • 28) 0.380 974 08 × 2 = 0 + 0.761 948 16;
  • 29) 0.761 948 16 × 2 = 1 + 0.523 896 32;
  • 30) 0.523 896 32 × 2 = 1 + 0.047 792 64;
  • 31) 0.047 792 64 × 2 = 0 + 0.095 585 28;
  • 32) 0.095 585 28 × 2 = 0 + 0.191 170 56;
  • 33) 0.191 170 56 × 2 = 0 + 0.382 341 12;
  • 34) 0.382 341 12 × 2 = 0 + 0.764 682 24;
  • 35) 0.764 682 24 × 2 = 1 + 0.529 364 48;
  • 36) 0.529 364 48 × 2 = 1 + 0.058 728 96;
  • 37) 0.058 728 96 × 2 = 0 + 0.117 457 92;
  • 38) 0.117 457 92 × 2 = 0 + 0.234 915 84;
  • 39) 0.234 915 84 × 2 = 0 + 0.469 831 68;
  • 40) 0.469 831 68 × 2 = 0 + 0.939 663 36;
  • 41) 0.939 663 36 × 2 = 1 + 0.879 326 72;
  • 42) 0.879 326 72 × 2 = 1 + 0.758 653 44;
  • 43) 0.758 653 44 × 2 = 1 + 0.517 306 88;
  • 44) 0.517 306 88 × 2 = 1 + 0.034 613 76;
  • 45) 0.034 613 76 × 2 = 0 + 0.069 227 52;
  • 46) 0.069 227 52 × 2 = 0 + 0.138 455 04;
  • 47) 0.138 455 04 × 2 = 0 + 0.276 910 08;
  • 48) 0.276 910 08 × 2 = 0 + 0.553 820 16;
  • 49) 0.553 820 16 × 2 = 1 + 0.107 640 32;
  • 50) 0.107 640 32 × 2 = 0 + 0.215 280 64;
  • 51) 0.215 280 64 × 2 = 0 + 0.430 561 28;
  • 52) 0.430 561 28 × 2 = 0 + 0.861 122 56;
  • 53) 0.861 122 56 × 2 = 1 + 0.722 245 12;
  • 54) 0.722 245 12 × 2 = 1 + 0.444 490 24;
  • 55) 0.444 490 24 × 2 = 0 + 0.888 980 48;
  • 56) 0.888 980 48 × 2 = 1 + 0.777 960 96;
  • 57) 0.777 960 96 × 2 = 1 + 0.555 921 92;
  • 58) 0.555 921 92 × 2 = 1 + 0.111 843 84;
  • 59) 0.111 843 84 × 2 = 0 + 0.223 687 68;
  • 60) 0.223 687 68 × 2 = 0 + 0.447 375 36;
  • 61) 0.447 375 36 × 2 = 0 + 0.894 750 72;
  • 62) 0.894 750 72 × 2 = 1 + 0.789 501 44;
  • 63) 0.789 501 44 × 2 = 1 + 0.579 002 88;
  • 64) 0.579 002 88 × 2 = 1 + 0.158 005 76;

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:


0.000 283 11(10) =

0.0000 0000 0001 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111(2)

6. Positive number before normalization:

0.000 283 11(10) =

0.0000 0000 0001 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 12 positions to the right, so that only one non zero digit remains to the left of it:


0.000 283 11(10) =

0.0000 0000 0001 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111(2) =

0.0000 0000 0001 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111(2) × 20 =

1.0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111(2) × 2-12


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): -12

Mantissa (not normalized):
1.0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-12 + 2(11-1) - 1 =

(-12 + 1 023)(10) =

1 011(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;
  • 1 011 ÷ 2 = 505 + 1;
  • 505 ÷ 2 = 252 + 1;
  • 252 ÷ 2 = 126 + 0;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1011(10) =

011 1111 0011(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. 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111 =

0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111


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 0011

Mantissa (52 bits) =
0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111


Decimal number -0.000 283 11 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 1101 1100 1100 0011 0000 1111 0000 1000 1101 1100 0111

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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