-0.000 282 68 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 68(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
-0.000 282 68(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:

|-0.000 282 68| = 0.000 282 68


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: 0.000 282 68.

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) 0.000 282 68 × 2 = 0 + 0.000 565 36;
  • 2) 0.000 565 36 × 2 = 0 + 0.001 130 72;
  • 3) 0.001 130 72 × 2 = 0 + 0.002 261 44;
  • 4) 0.002 261 44 × 2 = 0 + 0.004 522 88;
  • 5) 0.004 522 88 × 2 = 0 + 0.009 045 76;
  • 6) 0.009 045 76 × 2 = 0 + 0.018 091 52;
  • 7) 0.018 091 52 × 2 = 0 + 0.036 183 04;
  • 8) 0.036 183 04 × 2 = 0 + 0.072 366 08;
  • 9) 0.072 366 08 × 2 = 0 + 0.144 732 16;
  • 10) 0.144 732 16 × 2 = 0 + 0.289 464 32;
  • 11) 0.289 464 32 × 2 = 0 + 0.578 928 64;
  • 12) 0.578 928 64 × 2 = 1 + 0.157 857 28;
  • 13) 0.157 857 28 × 2 = 0 + 0.315 714 56;
  • 14) 0.315 714 56 × 2 = 0 + 0.631 429 12;
  • 15) 0.631 429 12 × 2 = 1 + 0.262 858 24;
  • 16) 0.262 858 24 × 2 = 0 + 0.525 716 48;
  • 17) 0.525 716 48 × 2 = 1 + 0.051 432 96;
  • 18) 0.051 432 96 × 2 = 0 + 0.102 865 92;
  • 19) 0.102 865 92 × 2 = 0 + 0.205 731 84;
  • 20) 0.205 731 84 × 2 = 0 + 0.411 463 68;
  • 21) 0.411 463 68 × 2 = 0 + 0.822 927 36;
  • 22) 0.822 927 36 × 2 = 1 + 0.645 854 72;
  • 23) 0.645 854 72 × 2 = 1 + 0.291 709 44;
  • 24) 0.291 709 44 × 2 = 0 + 0.583 418 88;
  • 25) 0.583 418 88 × 2 = 1 + 0.166 837 76;
  • 26) 0.166 837 76 × 2 = 0 + 0.333 675 52;
  • 27) 0.333 675 52 × 2 = 0 + 0.667 351 04;
  • 28) 0.667 351 04 × 2 = 1 + 0.334 702 08;
  • 29) 0.334 702 08 × 2 = 0 + 0.669 404 16;
  • 30) 0.669 404 16 × 2 = 1 + 0.338 808 32;
  • 31) 0.338 808 32 × 2 = 0 + 0.677 616 64;
  • 32) 0.677 616 64 × 2 = 1 + 0.355 233 28;
  • 33) 0.355 233 28 × 2 = 0 + 0.710 466 56;
  • 34) 0.710 466 56 × 2 = 1 + 0.420 933 12;
  • 35) 0.420 933 12 × 2 = 0 + 0.841 866 24;
  • 36) 0.841 866 24 × 2 = 1 + 0.683 732 48;
  • 37) 0.683 732 48 × 2 = 1 + 0.367 464 96;
  • 38) 0.367 464 96 × 2 = 0 + 0.734 929 92;
  • 39) 0.734 929 92 × 2 = 1 + 0.469 859 84;
  • 40) 0.469 859 84 × 2 = 0 + 0.939 719 68;
  • 41) 0.939 719 68 × 2 = 1 + 0.879 439 36;
  • 42) 0.879 439 36 × 2 = 1 + 0.758 878 72;
  • 43) 0.758 878 72 × 2 = 1 + 0.517 757 44;
  • 44) 0.517 757 44 × 2 = 1 + 0.035 514 88;
  • 45) 0.035 514 88 × 2 = 0 + 0.071 029 76;
  • 46) 0.071 029 76 × 2 = 0 + 0.142 059 52;
  • 47) 0.142 059 52 × 2 = 0 + 0.284 119 04;
  • 48) 0.284 119 04 × 2 = 0 + 0.568 238 08;
  • 49) 0.568 238 08 × 2 = 1 + 0.136 476 16;
  • 50) 0.136 476 16 × 2 = 0 + 0.272 952 32;
  • 51) 0.272 952 32 × 2 = 0 + 0.545 904 64;
  • 52) 0.545 904 64 × 2 = 1 + 0.091 809 28;
  • 53) 0.091 809 28 × 2 = 0 + 0.183 618 56;
  • 54) 0.183 618 56 × 2 = 0 + 0.367 237 12;
  • 55) 0.367 237 12 × 2 = 0 + 0.734 474 24;
  • 56) 0.734 474 24 × 2 = 1 + 0.468 948 48;
  • 57) 0.468 948 48 × 2 = 0 + 0.937 896 96;
  • 58) 0.937 896 96 × 2 = 1 + 0.875 793 92;
  • 59) 0.875 793 92 × 2 = 1 + 0.751 587 84;
  • 60) 0.751 587 84 × 2 = 1 + 0.503 175 68;
  • 61) 0.503 175 68 × 2 = 1 + 0.006 351 36;
  • 62) 0.006 351 36 × 2 = 0 + 0.012 702 72;
  • 63) 0.012 702 72 × 2 = 0 + 0.025 405 44;
  • 64) 0.025 405 44 × 2 = 0 + 0.050 810 88;

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:


0.000 282 68(10) =

0.0000 0000 0001 0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000(2)

6. Positive number before normalization:

0.000 282 68(10) =

0.0000 0000 0001 0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000(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:


0.000 282 68(10) =

0.0000 0000 0001 0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000(2) =

0.0000 0000 0001 0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000(2) × 20 =

1.0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000(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 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000


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 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000 =

0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 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 1111 0011

Mantissa (52 bits) =
0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 1000


Decimal number -0.000 282 68 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0110 1001 0101 0101 1010 1111 0000 1001 0001 0111 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