-0.000 281 13 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 281 13(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
-0.000 281 13(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:

|-0.000 281 13| = 0.000 281 13


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: 0.000 281 13.

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) 0.000 281 13 × 2 = 0 + 0.000 562 26;
  • 2) 0.000 562 26 × 2 = 0 + 0.001 124 52;
  • 3) 0.001 124 52 × 2 = 0 + 0.002 249 04;
  • 4) 0.002 249 04 × 2 = 0 + 0.004 498 08;
  • 5) 0.004 498 08 × 2 = 0 + 0.008 996 16;
  • 6) 0.008 996 16 × 2 = 0 + 0.017 992 32;
  • 7) 0.017 992 32 × 2 = 0 + 0.035 984 64;
  • 8) 0.035 984 64 × 2 = 0 + 0.071 969 28;
  • 9) 0.071 969 28 × 2 = 0 + 0.143 938 56;
  • 10) 0.143 938 56 × 2 = 0 + 0.287 877 12;
  • 11) 0.287 877 12 × 2 = 0 + 0.575 754 24;
  • 12) 0.575 754 24 × 2 = 1 + 0.151 508 48;
  • 13) 0.151 508 48 × 2 = 0 + 0.303 016 96;
  • 14) 0.303 016 96 × 2 = 0 + 0.606 033 92;
  • 15) 0.606 033 92 × 2 = 1 + 0.212 067 84;
  • 16) 0.212 067 84 × 2 = 0 + 0.424 135 68;
  • 17) 0.424 135 68 × 2 = 0 + 0.848 271 36;
  • 18) 0.848 271 36 × 2 = 1 + 0.696 542 72;
  • 19) 0.696 542 72 × 2 = 1 + 0.393 085 44;
  • 20) 0.393 085 44 × 2 = 0 + 0.786 170 88;
  • 21) 0.786 170 88 × 2 = 1 + 0.572 341 76;
  • 22) 0.572 341 76 × 2 = 1 + 0.144 683 52;
  • 23) 0.144 683 52 × 2 = 0 + 0.289 367 04;
  • 24) 0.289 367 04 × 2 = 0 + 0.578 734 08;
  • 25) 0.578 734 08 × 2 = 1 + 0.157 468 16;
  • 26) 0.157 468 16 × 2 = 0 + 0.314 936 32;
  • 27) 0.314 936 32 × 2 = 0 + 0.629 872 64;
  • 28) 0.629 872 64 × 2 = 1 + 0.259 745 28;
  • 29) 0.259 745 28 × 2 = 0 + 0.519 490 56;
  • 30) 0.519 490 56 × 2 = 1 + 0.038 981 12;
  • 31) 0.038 981 12 × 2 = 0 + 0.077 962 24;
  • 32) 0.077 962 24 × 2 = 0 + 0.155 924 48;
  • 33) 0.155 924 48 × 2 = 0 + 0.311 848 96;
  • 34) 0.311 848 96 × 2 = 0 + 0.623 697 92;
  • 35) 0.623 697 92 × 2 = 1 + 0.247 395 84;
  • 36) 0.247 395 84 × 2 = 0 + 0.494 791 68;
  • 37) 0.494 791 68 × 2 = 0 + 0.989 583 36;
  • 38) 0.989 583 36 × 2 = 1 + 0.979 166 72;
  • 39) 0.979 166 72 × 2 = 1 + 0.958 333 44;
  • 40) 0.958 333 44 × 2 = 1 + 0.916 666 88;
  • 41) 0.916 666 88 × 2 = 1 + 0.833 333 76;
  • 42) 0.833 333 76 × 2 = 1 + 0.666 667 52;
  • 43) 0.666 667 52 × 2 = 1 + 0.333 335 04;
  • 44) 0.333 335 04 × 2 = 0 + 0.666 670 08;
  • 45) 0.666 670 08 × 2 = 1 + 0.333 340 16;
  • 46) 0.333 340 16 × 2 = 0 + 0.666 680 32;
  • 47) 0.666 680 32 × 2 = 1 + 0.333 360 64;
  • 48) 0.333 360 64 × 2 = 0 + 0.666 721 28;
  • 49) 0.666 721 28 × 2 = 1 + 0.333 442 56;
  • 50) 0.333 442 56 × 2 = 0 + 0.666 885 12;
  • 51) 0.666 885 12 × 2 = 1 + 0.333 770 24;
  • 52) 0.333 770 24 × 2 = 0 + 0.667 540 48;
  • 53) 0.667 540 48 × 2 = 1 + 0.335 080 96;
  • 54) 0.335 080 96 × 2 = 0 + 0.670 161 92;
  • 55) 0.670 161 92 × 2 = 1 + 0.340 323 84;
  • 56) 0.340 323 84 × 2 = 0 + 0.680 647 68;
  • 57) 0.680 647 68 × 2 = 1 + 0.361 295 36;
  • 58) 0.361 295 36 × 2 = 0 + 0.722 590 72;
  • 59) 0.722 590 72 × 2 = 1 + 0.445 181 44;
  • 60) 0.445 181 44 × 2 = 0 + 0.890 362 88;
  • 61) 0.890 362 88 × 2 = 1 + 0.780 725 76;
  • 62) 0.780 725 76 × 2 = 1 + 0.561 451 52;
  • 63) 0.561 451 52 × 2 = 1 + 0.122 903 04;
  • 64) 0.122 903 04 × 2 = 0 + 0.245 806 08;

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:


0.000 281 13(10) =

0.0000 0000 0001 0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110(2)

6. Positive number before normalization:

0.000 281 13(10) =

0.0000 0000 0001 0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110(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:


0.000 281 13(10) =

0.0000 0000 0001 0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110(2) =

0.0000 0000 0001 0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110(2) × 20 =

1.0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110(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 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110


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 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110 =

0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110


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 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110


Decimal number -0.000 281 13 converted to 64 bit double precision IEEE 754 binary floating point representation:

1 - 011 1111 0011 - 0010 0110 1100 1001 0100 0010 0111 1110 1010 1010 1010 1010 1110

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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