Convert 10.103 985 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

10.103 985(10) to 64 bit double precision IEEE 754 binary floating point (1 bit for sign, 11 bits for exponent, 52 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 10.
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;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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.


10(10) =

1010(2)


3. Convert to the binary (base 2) the fractional part: 0.103 985.

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) 0.103 985 × 2 = 0 + 0.207 97;
  • 2) 0.207 97 × 2 = 0 + 0.415 94;
  • 3) 0.415 94 × 2 = 0 + 0.831 88;
  • 4) 0.831 88 × 2 = 1 + 0.663 76;
  • 5) 0.663 76 × 2 = 1 + 0.327 52;
  • 6) 0.327 52 × 2 = 0 + 0.655 04;
  • 7) 0.655 04 × 2 = 1 + 0.310 08;
  • 8) 0.310 08 × 2 = 0 + 0.620 16;
  • 9) 0.620 16 × 2 = 1 + 0.240 32;
  • 10) 0.240 32 × 2 = 0 + 0.480 64;
  • 11) 0.480 64 × 2 = 0 + 0.961 28;
  • 12) 0.961 28 × 2 = 1 + 0.922 56;
  • 13) 0.922 56 × 2 = 1 + 0.845 12;
  • 14) 0.845 12 × 2 = 1 + 0.690 24;
  • 15) 0.690 24 × 2 = 1 + 0.380 48;
  • 16) 0.380 48 × 2 = 0 + 0.760 96;
  • 17) 0.760 96 × 2 = 1 + 0.521 92;
  • 18) 0.521 92 × 2 = 1 + 0.043 84;
  • 19) 0.043 84 × 2 = 0 + 0.087 68;
  • 20) 0.087 68 × 2 = 0 + 0.175 36;
  • 21) 0.175 36 × 2 = 0 + 0.350 72;
  • 22) 0.350 72 × 2 = 0 + 0.701 44;
  • 23) 0.701 44 × 2 = 1 + 0.402 88;
  • 24) 0.402 88 × 2 = 0 + 0.805 76;
  • 25) 0.805 76 × 2 = 1 + 0.611 52;
  • 26) 0.611 52 × 2 = 1 + 0.223 04;
  • 27) 0.223 04 × 2 = 0 + 0.446 08;
  • 28) 0.446 08 × 2 = 0 + 0.892 16;
  • 29) 0.892 16 × 2 = 1 + 0.784 32;
  • 30) 0.784 32 × 2 = 1 + 0.568 64;
  • 31) 0.568 64 × 2 = 1 + 0.137 28;
  • 32) 0.137 28 × 2 = 0 + 0.274 56;
  • 33) 0.274 56 × 2 = 0 + 0.549 12;
  • 34) 0.549 12 × 2 = 1 + 0.098 24;
  • 35) 0.098 24 × 2 = 0 + 0.196 48;
  • 36) 0.196 48 × 2 = 0 + 0.392 96;
  • 37) 0.392 96 × 2 = 0 + 0.785 92;
  • 38) 0.785 92 × 2 = 1 + 0.571 84;
  • 39) 0.571 84 × 2 = 1 + 0.143 68;
  • 40) 0.143 68 × 2 = 0 + 0.287 36;
  • 41) 0.287 36 × 2 = 0 + 0.574 72;
  • 42) 0.574 72 × 2 = 1 + 0.149 44;
  • 43) 0.149 44 × 2 = 0 + 0.298 88;
  • 44) 0.298 88 × 2 = 0 + 0.597 76;
  • 45) 0.597 76 × 2 = 1 + 0.195 52;
  • 46) 0.195 52 × 2 = 0 + 0.391 04;
  • 47) 0.391 04 × 2 = 0 + 0.782 08;
  • 48) 0.782 08 × 2 = 1 + 0.564 16;
  • 49) 0.564 16 × 2 = 1 + 0.128 32;
  • 50) 0.128 32 × 2 = 0 + 0.256 64;
  • 51) 0.256 64 × 2 = 0 + 0.513 28;
  • 52) 0.513 28 × 2 = 1 + 0.026 56;
  • 53) 0.026 56 × 2 = 0 + 0.053 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...)


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:


0.103 985(10) =

0.0001 1010 1001 1110 1100 0010 1100 1110 0100 0110 0100 1001 1001 0(2)


5. Positive number before normalization:

10.103 985(10) =

1010.0001 1010 1001 1110 1100 0010 1100 1110 0100 0110 0100 1001 1001 0(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left so that only one non zero digit remains to the left of it:


10.103 985(10) =

1010.0001 1010 1001 1110 1100 0010 1100 1110 0100 0110 0100 1001 1001 0(2) =

1010.0001 1010 1001 1110 1100 0010 1100 1110 0100 0110 0100 1001 1001 0(2) × 20 =

1.0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011 0010(2) × 23


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): 3

Mantissa (not normalized):
1.0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

3 + 2(11-1) - 1 =

(3 + 1 023)(10) =

1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

1026(10) =

100 0000 0010(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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011 0010 =

0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011


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) =
100 0000 0010

Mantissa (52 bits) =
0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011


Number 10.103 985 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 100 0000 0010 - 0100 0011 0101 0011 1101 1000 0101 1001 1100 1000 1100 1001 0011

(64 bits IEEE 754)

More operations of this kind:

10.103 984 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point = ?

10.103 986 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point = ?


Convert to 64 bit double precision IEEE 754 binary floating point standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes one bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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All base ten decimal numbers converted to 64 bit double precision IEEE 754 binary floating point

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