Convert 0.108 71 to 64 Bit Double Precision IEEE 754 Binary Floating Point Standard, From a Number in Base 10 Decimal System

0.108 71(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: 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 the binary (base 2) the fractional part: 0.108 71.

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) 0.108 71 × 2 = 0 + 0.217 42;
  • 2) 0.217 42 × 2 = 0 + 0.434 84;
  • 3) 0.434 84 × 2 = 0 + 0.869 68;
  • 4) 0.869 68 × 2 = 1 + 0.739 36;
  • 5) 0.739 36 × 2 = 1 + 0.478 72;
  • 6) 0.478 72 × 2 = 0 + 0.957 44;
  • 7) 0.957 44 × 2 = 1 + 0.914 88;
  • 8) 0.914 88 × 2 = 1 + 0.829 76;
  • 9) 0.829 76 × 2 = 1 + 0.659 52;
  • 10) 0.659 52 × 2 = 1 + 0.319 04;
  • 11) 0.319 04 × 2 = 0 + 0.638 08;
  • 12) 0.638 08 × 2 = 1 + 0.276 16;
  • 13) 0.276 16 × 2 = 0 + 0.552 32;
  • 14) 0.552 32 × 2 = 1 + 0.104 64;
  • 15) 0.104 64 × 2 = 0 + 0.209 28;
  • 16) 0.209 28 × 2 = 0 + 0.418 56;
  • 17) 0.418 56 × 2 = 0 + 0.837 12;
  • 18) 0.837 12 × 2 = 1 + 0.674 24;
  • 19) 0.674 24 × 2 = 1 + 0.348 48;
  • 20) 0.348 48 × 2 = 0 + 0.696 96;
  • 21) 0.696 96 × 2 = 1 + 0.393 92;
  • 22) 0.393 92 × 2 = 0 + 0.787 84;
  • 23) 0.787 84 × 2 = 1 + 0.575 68;
  • 24) 0.575 68 × 2 = 1 + 0.151 36;
  • 25) 0.151 36 × 2 = 0 + 0.302 72;
  • 26) 0.302 72 × 2 = 0 + 0.605 44;
  • 27) 0.605 44 × 2 = 1 + 0.210 88;
  • 28) 0.210 88 × 2 = 0 + 0.421 76;
  • 29) 0.421 76 × 2 = 0 + 0.843 52;
  • 30) 0.843 52 × 2 = 1 + 0.687 04;
  • 31) 0.687 04 × 2 = 1 + 0.374 08;
  • 32) 0.374 08 × 2 = 0 + 0.748 16;
  • 33) 0.748 16 × 2 = 1 + 0.496 32;
  • 34) 0.496 32 × 2 = 0 + 0.992 64;
  • 35) 0.992 64 × 2 = 1 + 0.985 28;
  • 36) 0.985 28 × 2 = 1 + 0.970 56;
  • 37) 0.970 56 × 2 = 1 + 0.941 12;
  • 38) 0.941 12 × 2 = 1 + 0.882 24;
  • 39) 0.882 24 × 2 = 1 + 0.764 48;
  • 40) 0.764 48 × 2 = 1 + 0.528 96;
  • 41) 0.528 96 × 2 = 1 + 0.057 92;
  • 42) 0.057 92 × 2 = 0 + 0.115 84;
  • 43) 0.115 84 × 2 = 0 + 0.231 68;
  • 44) 0.231 68 × 2 = 0 + 0.463 36;
  • 45) 0.463 36 × 2 = 0 + 0.926 72;
  • 46) 0.926 72 × 2 = 1 + 0.853 44;
  • 47) 0.853 44 × 2 = 1 + 0.706 88;
  • 48) 0.706 88 × 2 = 1 + 0.413 76;
  • 49) 0.413 76 × 2 = 0 + 0.827 52;
  • 50) 0.827 52 × 2 = 1 + 0.655 04;
  • 51) 0.655 04 × 2 = 1 + 0.310 08;
  • 52) 0.310 08 × 2 = 0 + 0.620 16;
  • 53) 0.620 16 × 2 = 1 + 0.240 32;
  • 54) 0.240 32 × 2 = 0 + 0.480 64;
  • 55) 0.480 64 × 2 = 0 + 0.961 28;
  • 56) 0.961 28 × 2 = 1 + 0.922 56;

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:

0.108 71(10) =


0.0001 1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001(2)


5. Positive number before normalization:

0.108 71(10) =


0.0001 1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right so that only one non zero digit remains to the left of it:

0.108 71(10) =


0.0001 1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001(2) =


0.0001 1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001(2) × 20 =


1.1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001(2) × 2-4


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): -4


Mantissa (not normalized):
1.1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:

Exponent (adjusted) =


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


-4 + 2(11-1) - 1 =


(-4 + 1 023)(10) =


1 019(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;
  • 1 019 ÷ 2 = 509 + 1;
  • 509 ÷ 2 = 254 + 1;
  • 254 ÷ 2 = 127 + 0;
  • 127 ÷ 2 = 63 + 1;
  • 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) =


1019(10) =


011 1111 1011(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 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001 =


1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 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 1011


Mantissa (52 bits) =
1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001


Number 0.108 71 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point:
0 - 011 1111 1011 - 1011 1101 0100 0110 1011 0010 0110 1011 1111 1000 0111 0110 1001

(64 bits IEEE 754)
  • Sign (1 bit):

    • 0

      63
  • Exponent (11 bits):

    • 0

      62
    • 1

      61
    • 1

      60
    • 1

      59
    • 1

      58
    • 1

      57
    • 1

      56
    • 1

      55
    • 0

      54
    • 1

      53
    • 1

      52
  • Mantissa (52 bits):

    • 1

      51
    • 0

      50
    • 1

      49
    • 1

      48
    • 1

      47
    • 1

      46
    • 0

      45
    • 1

      44
    • 0

      43
    • 1

      42
    • 0

      41
    • 0

      40
    • 0

      39
    • 1

      38
    • 1

      37
    • 0

      36
    • 1

      35
    • 0

      34
    • 1

      33
    • 1

      32
    • 0

      31
    • 0

      30
    • 1

      29
    • 0

      28
    • 0

      27
    • 1

      26
    • 1

      25
    • 0

      24
    • 1

      23
    • 0

      22
    • 1

      21
    • 1

      20
    • 1

      19
    • 1

      18
    • 1

      17
    • 1

      16
    • 1

      15
    • 0

      14
    • 0

      13
    • 0

      12
    • 0

      11
    • 1

      10
    • 1

      9
    • 1

      8
    • 0

      7
    • 1

      6
    • 1

      5
    • 0

      4
    • 1

      3
    • 0

      2
    • 0

      1
    • 1

      0

More operations of this kind:

0.108 7 = ? ... 0.108 72 = ?


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)

Latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation

0.108 71 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:37 UTC (GMT)
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1.333 34 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
0.190 3 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
27.33 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
20.5 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
0.000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 4 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
-13.1 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
2.381 918 1 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:36 UTC (GMT)
0.785 398 163 458 219 169 011 004 073 672 148 207 311 201 868 187 2 to 64 bit double precision IEEE 754 binary floating point = ? Mar 08 12:35 UTC (GMT)
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