64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 0.006 68 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 0.006 68(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.006 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) 0.006 68 × 2 = 0 + 0.013 36;
  • 2) 0.013 36 × 2 = 0 + 0.026 72;
  • 3) 0.026 72 × 2 = 0 + 0.053 44;
  • 4) 0.053 44 × 2 = 0 + 0.106 88;
  • 5) 0.106 88 × 2 = 0 + 0.213 76;
  • 6) 0.213 76 × 2 = 0 + 0.427 52;
  • 7) 0.427 52 × 2 = 0 + 0.855 04;
  • 8) 0.855 04 × 2 = 1 + 0.710 08;
  • 9) 0.710 08 × 2 = 1 + 0.420 16;
  • 10) 0.420 16 × 2 = 0 + 0.840 32;
  • 11) 0.840 32 × 2 = 1 + 0.680 64;
  • 12) 0.680 64 × 2 = 1 + 0.361 28;
  • 13) 0.361 28 × 2 = 0 + 0.722 56;
  • 14) 0.722 56 × 2 = 1 + 0.445 12;
  • 15) 0.445 12 × 2 = 0 + 0.890 24;
  • 16) 0.890 24 × 2 = 1 + 0.780 48;
  • 17) 0.780 48 × 2 = 1 + 0.560 96;
  • 18) 0.560 96 × 2 = 1 + 0.121 92;
  • 19) 0.121 92 × 2 = 0 + 0.243 84;
  • 20) 0.243 84 × 2 = 0 + 0.487 68;
  • 21) 0.487 68 × 2 = 0 + 0.975 36;
  • 22) 0.975 36 × 2 = 1 + 0.950 72;
  • 23) 0.950 72 × 2 = 1 + 0.901 44;
  • 24) 0.901 44 × 2 = 1 + 0.802 88;
  • 25) 0.802 88 × 2 = 1 + 0.605 76;
  • 26) 0.605 76 × 2 = 1 + 0.211 52;
  • 27) 0.211 52 × 2 = 0 + 0.423 04;
  • 28) 0.423 04 × 2 = 0 + 0.846 08;
  • 29) 0.846 08 × 2 = 1 + 0.692 16;
  • 30) 0.692 16 × 2 = 1 + 0.384 32;
  • 31) 0.384 32 × 2 = 0 + 0.768 64;
  • 32) 0.768 64 × 2 = 1 + 0.537 28;
  • 33) 0.537 28 × 2 = 1 + 0.074 56;
  • 34) 0.074 56 × 2 = 0 + 0.149 12;
  • 35) 0.149 12 × 2 = 0 + 0.298 24;
  • 36) 0.298 24 × 2 = 0 + 0.596 48;
  • 37) 0.596 48 × 2 = 1 + 0.192 96;
  • 38) 0.192 96 × 2 = 0 + 0.385 92;
  • 39) 0.385 92 × 2 = 0 + 0.771 84;
  • 40) 0.771 84 × 2 = 1 + 0.543 68;
  • 41) 0.543 68 × 2 = 1 + 0.087 36;
  • 42) 0.087 36 × 2 = 0 + 0.174 72;
  • 43) 0.174 72 × 2 = 0 + 0.349 44;
  • 44) 0.349 44 × 2 = 0 + 0.698 88;
  • 45) 0.698 88 × 2 = 1 + 0.397 76;
  • 46) 0.397 76 × 2 = 0 + 0.795 52;
  • 47) 0.795 52 × 2 = 1 + 0.591 04;
  • 48) 0.591 04 × 2 = 1 + 0.182 08;
  • 49) 0.182 08 × 2 = 0 + 0.364 16;
  • 50) 0.364 16 × 2 = 0 + 0.728 32;
  • 51) 0.728 32 × 2 = 1 + 0.456 64;
  • 52) 0.456 64 × 2 = 0 + 0.913 28;
  • 53) 0.913 28 × 2 = 1 + 0.826 56;
  • 54) 0.826 56 × 2 = 1 + 0.653 12;
  • 55) 0.653 12 × 2 = 1 + 0.306 24;
  • 56) 0.306 24 × 2 = 0 + 0.612 48;
  • 57) 0.612 48 × 2 = 1 + 0.224 96;
  • 58) 0.224 96 × 2 = 0 + 0.449 92;
  • 59) 0.449 92 × 2 = 0 + 0.899 84;
  • 60) 0.899 84 × 2 = 1 + 0.799 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...)


4. 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:


0.006 68(10) =

0.0000 0001 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001(2)


5. Positive number before normalization:

0.006 68(10) =

0.0000 0001 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 8 positions to the right, so that only one non zero digit remains to the left of it:


0.006 68(10) =

0.0000 0001 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001(2) =

0.0000 0001 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001(2) × 20 =

1.1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001(2) × 2-8


7. 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 0 (a positive number)

Exponent (unadjusted): -8

Mantissa (not normalized):
1.1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

-8 + 2(11-1) - 1 =

(-8 + 1 023)(10) =

1 015(10)


9. 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;
  • 1 015 ÷ 2 = 507 + 1;
  • 507 ÷ 2 = 253 + 1;
  • 253 ÷ 2 = 126 + 1;
  • 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;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

1015(10) =

011 1111 0111(2)


11. 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. 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001 =

1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1111 0111

Mantissa (52 bits) =
1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001


The base ten decimal number 0.006 68 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 011 1111 0111 - 1011 0101 1100 0111 1100 1101 1000 1001 1000 1011 0010 1110 1001

More operations with decimal numbers converted to 64 bit double precision IEEE 754 binary floating point representation:

» Number 0.006 67 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

» Number 0.006 69 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

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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