0.110 099 998 64 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.110 099 998 64(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
0.110 099 998 64(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. 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: 0.110 099 998 64.

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) 0.110 099 998 64 × 2 = 0 + 0.220 199 997 28;
  • 2) 0.220 199 997 28 × 2 = 0 + 0.440 399 994 56;
  • 3) 0.440 399 994 56 × 2 = 0 + 0.880 799 989 12;
  • 4) 0.880 799 989 12 × 2 = 1 + 0.761 599 978 24;
  • 5) 0.761 599 978 24 × 2 = 1 + 0.523 199 956 48;
  • 6) 0.523 199 956 48 × 2 = 1 + 0.046 399 912 96;
  • 7) 0.046 399 912 96 × 2 = 0 + 0.092 799 825 92;
  • 8) 0.092 799 825 92 × 2 = 0 + 0.185 599 651 84;
  • 9) 0.185 599 651 84 × 2 = 0 + 0.371 199 303 68;
  • 10) 0.371 199 303 68 × 2 = 0 + 0.742 398 607 36;
  • 11) 0.742 398 607 36 × 2 = 1 + 0.484 797 214 72;
  • 12) 0.484 797 214 72 × 2 = 0 + 0.969 594 429 44;
  • 13) 0.969 594 429 44 × 2 = 1 + 0.939 188 858 88;
  • 14) 0.939 188 858 88 × 2 = 1 + 0.878 377 717 76;
  • 15) 0.878 377 717 76 × 2 = 1 + 0.756 755 435 52;
  • 16) 0.756 755 435 52 × 2 = 1 + 0.513 510 871 04;
  • 17) 0.513 510 871 04 × 2 = 1 + 0.027 021 742 08;
  • 18) 0.027 021 742 08 × 2 = 0 + 0.054 043 484 16;
  • 19) 0.054 043 484 16 × 2 = 0 + 0.108 086 968 32;
  • 20) 0.108 086 968 32 × 2 = 0 + 0.216 173 936 64;
  • 21) 0.216 173 936 64 × 2 = 0 + 0.432 347 873 28;
  • 22) 0.432 347 873 28 × 2 = 0 + 0.864 695 746 56;
  • 23) 0.864 695 746 56 × 2 = 1 + 0.729 391 493 12;
  • 24) 0.729 391 493 12 × 2 = 1 + 0.458 782 986 24;
  • 25) 0.458 782 986 24 × 2 = 0 + 0.917 565 972 48;
  • 26) 0.917 565 972 48 × 2 = 1 + 0.835 131 944 96;
  • 27) 0.835 131 944 96 × 2 = 1 + 0.670 263 889 92;
  • 28) 0.670 263 889 92 × 2 = 1 + 0.340 527 779 84;
  • 29) 0.340 527 779 84 × 2 = 0 + 0.681 055 559 68;
  • 30) 0.681 055 559 68 × 2 = 1 + 0.362 111 119 36;
  • 31) 0.362 111 119 36 × 2 = 0 + 0.724 222 238 72;
  • 32) 0.724 222 238 72 × 2 = 1 + 0.448 444 477 44;
  • 33) 0.448 444 477 44 × 2 = 0 + 0.896 888 954 88;
  • 34) 0.896 888 954 88 × 2 = 1 + 0.793 777 909 76;
  • 35) 0.793 777 909 76 × 2 = 1 + 0.587 555 819 52;
  • 36) 0.587 555 819 52 × 2 = 1 + 0.175 111 639 04;
  • 37) 0.175 111 639 04 × 2 = 0 + 0.350 223 278 08;
  • 38) 0.350 223 278 08 × 2 = 0 + 0.700 446 556 16;
  • 39) 0.700 446 556 16 × 2 = 1 + 0.400 893 112 32;
  • 40) 0.400 893 112 32 × 2 = 0 + 0.801 786 224 64;
  • 41) 0.801 786 224 64 × 2 = 1 + 0.603 572 449 28;
  • 42) 0.603 572 449 28 × 2 = 1 + 0.207 144 898 56;
  • 43) 0.207 144 898 56 × 2 = 0 + 0.414 289 797 12;
  • 44) 0.414 289 797 12 × 2 = 0 + 0.828 579 594 24;
  • 45) 0.828 579 594 24 × 2 = 1 + 0.657 159 188 48;
  • 46) 0.657 159 188 48 × 2 = 1 + 0.314 318 376 96;
  • 47) 0.314 318 376 96 × 2 = 0 + 0.628 636 753 92;
  • 48) 0.628 636 753 92 × 2 = 1 + 0.257 273 507 84;
  • 49) 0.257 273 507 84 × 2 = 0 + 0.514 547 015 68;
  • 50) 0.514 547 015 68 × 2 = 1 + 0.029 094 031 36;
  • 51) 0.029 094 031 36 × 2 = 0 + 0.058 188 062 72;
  • 52) 0.058 188 062 72 × 2 = 0 + 0.116 376 125 44;
  • 53) 0.116 376 125 44 × 2 = 0 + 0.232 752 250 88;
  • 54) 0.232 752 250 88 × 2 = 0 + 0.465 504 501 76;
  • 55) 0.465 504 501 76 × 2 = 0 + 0.931 009 003 52;
  • 56) 0.931 009 003 52 × 2 = 1 + 0.862 018 007 04;

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).


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:


0.110 099 998 64(10) =

0.0001 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001(2)

5. Positive number before normalization:

0.110 099 998 64(10) =

0.0001 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001(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:


0.110 099 998 64(10) =

0.0001 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001(2) =

0.0001 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001(2) × 20 =

1.1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001(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):
1.1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001


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. 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001 =

1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001


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) =
1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001


Decimal number 0.110 099 998 64 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1100 0010 1111 1000 0011 0111 0101 0111 0010 1100 1101 0100 0001

» Convert decimal 0.110 099 999 32 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 10, 2026 [October]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: October - by our visitors

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