-0.000 282 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 282 32(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 32(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 32| = 0.000 282 32


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 32.

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 32 × 2 = 0 + 0.000 564 64;
  • 2) 0.000 564 64 × 2 = 0 + 0.001 129 28;
  • 3) 0.001 129 28 × 2 = 0 + 0.002 258 56;
  • 4) 0.002 258 56 × 2 = 0 + 0.004 517 12;
  • 5) 0.004 517 12 × 2 = 0 + 0.009 034 24;
  • 6) 0.009 034 24 × 2 = 0 + 0.018 068 48;
  • 7) 0.018 068 48 × 2 = 0 + 0.036 136 96;
  • 8) 0.036 136 96 × 2 = 0 + 0.072 273 92;
  • 9) 0.072 273 92 × 2 = 0 + 0.144 547 84;
  • 10) 0.144 547 84 × 2 = 0 + 0.289 095 68;
  • 11) 0.289 095 68 × 2 = 0 + 0.578 191 36;
  • 12) 0.578 191 36 × 2 = 1 + 0.156 382 72;
  • 13) 0.156 382 72 × 2 = 0 + 0.312 765 44;
  • 14) 0.312 765 44 × 2 = 0 + 0.625 530 88;
  • 15) 0.625 530 88 × 2 = 1 + 0.251 061 76;
  • 16) 0.251 061 76 × 2 = 0 + 0.502 123 52;
  • 17) 0.502 123 52 × 2 = 1 + 0.004 247 04;
  • 18) 0.004 247 04 × 2 = 0 + 0.008 494 08;
  • 19) 0.008 494 08 × 2 = 0 + 0.016 988 16;
  • 20) 0.016 988 16 × 2 = 0 + 0.033 976 32;
  • 21) 0.033 976 32 × 2 = 0 + 0.067 952 64;
  • 22) 0.067 952 64 × 2 = 0 + 0.135 905 28;
  • 23) 0.135 905 28 × 2 = 0 + 0.271 810 56;
  • 24) 0.271 810 56 × 2 = 0 + 0.543 621 12;
  • 25) 0.543 621 12 × 2 = 1 + 0.087 242 24;
  • 26) 0.087 242 24 × 2 = 0 + 0.174 484 48;
  • 27) 0.174 484 48 × 2 = 0 + 0.348 968 96;
  • 28) 0.348 968 96 × 2 = 0 + 0.697 937 92;
  • 29) 0.697 937 92 × 2 = 1 + 0.395 875 84;
  • 30) 0.395 875 84 × 2 = 0 + 0.791 751 68;
  • 31) 0.791 751 68 × 2 = 1 + 0.583 503 36;
  • 32) 0.583 503 36 × 2 = 1 + 0.167 006 72;
  • 33) 0.167 006 72 × 2 = 0 + 0.334 013 44;
  • 34) 0.334 013 44 × 2 = 0 + 0.668 026 88;
  • 35) 0.668 026 88 × 2 = 1 + 0.336 053 76;
  • 36) 0.336 053 76 × 2 = 0 + 0.672 107 52;
  • 37) 0.672 107 52 × 2 = 1 + 0.344 215 04;
  • 38) 0.344 215 04 × 2 = 0 + 0.688 430 08;
  • 39) 0.688 430 08 × 2 = 1 + 0.376 860 16;
  • 40) 0.376 860 16 × 2 = 0 + 0.753 720 32;
  • 41) 0.753 720 32 × 2 = 1 + 0.507 440 64;
  • 42) 0.507 440 64 × 2 = 1 + 0.014 881 28;
  • 43) 0.014 881 28 × 2 = 0 + 0.029 762 56;
  • 44) 0.029 762 56 × 2 = 0 + 0.059 525 12;
  • 45) 0.059 525 12 × 2 = 0 + 0.119 050 24;
  • 46) 0.119 050 24 × 2 = 0 + 0.238 100 48;
  • 47) 0.238 100 48 × 2 = 0 + 0.476 200 96;
  • 48) 0.476 200 96 × 2 = 0 + 0.952 401 92;
  • 49) 0.952 401 92 × 2 = 1 + 0.904 803 84;
  • 50) 0.904 803 84 × 2 = 1 + 0.809 607 68;
  • 51) 0.809 607 68 × 2 = 1 + 0.619 215 36;
  • 52) 0.619 215 36 × 2 = 1 + 0.238 430 72;
  • 53) 0.238 430 72 × 2 = 0 + 0.476 861 44;
  • 54) 0.476 861 44 × 2 = 0 + 0.953 722 88;
  • 55) 0.953 722 88 × 2 = 1 + 0.907 445 76;
  • 56) 0.907 445 76 × 2 = 1 + 0.814 891 52;
  • 57) 0.814 891 52 × 2 = 1 + 0.629 783 04;
  • 58) 0.629 783 04 × 2 = 1 + 0.259 566 08;
  • 59) 0.259 566 08 × 2 = 0 + 0.519 132 16;
  • 60) 0.519 132 16 × 2 = 1 + 0.038 264 32;
  • 61) 0.038 264 32 × 2 = 0 + 0.076 528 64;
  • 62) 0.076 528 64 × 2 = 0 + 0.153 057 28;
  • 63) 0.153 057 28 × 2 = 0 + 0.306 114 56;
  • 64) 0.306 114 56 × 2 = 0 + 0.612 229 12;

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 32(10) =

0.0000 0000 0001 0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000(2)

6. Positive number before normalization:

0.000 282 32(10) =

0.0000 0000 0001 0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000(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 32(10) =

0.0000 0000 0001 0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000(2) =

0.0000 0000 0001 0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000(2) × 20 =

1.0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000(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 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000


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 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000 =

0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 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 0011

Mantissa (52 bits) =
0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 0000


Decimal number -0.000 282 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 1000 0000 1000 1011 0010 1010 1100 0000 1111 0011 1101 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