0.110 099 998 26 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.110 099 998 26(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 26(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 26.

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 26 × 2 = 0 + 0.220 199 996 52;
  • 2) 0.220 199 996 52 × 2 = 0 + 0.440 399 993 04;
  • 3) 0.440 399 993 04 × 2 = 0 + 0.880 799 986 08;
  • 4) 0.880 799 986 08 × 2 = 1 + 0.761 599 972 16;
  • 5) 0.761 599 972 16 × 2 = 1 + 0.523 199 944 32;
  • 6) 0.523 199 944 32 × 2 = 1 + 0.046 399 888 64;
  • 7) 0.046 399 888 64 × 2 = 0 + 0.092 799 777 28;
  • 8) 0.092 799 777 28 × 2 = 0 + 0.185 599 554 56;
  • 9) 0.185 599 554 56 × 2 = 0 + 0.371 199 109 12;
  • 10) 0.371 199 109 12 × 2 = 0 + 0.742 398 218 24;
  • 11) 0.742 398 218 24 × 2 = 1 + 0.484 796 436 48;
  • 12) 0.484 796 436 48 × 2 = 0 + 0.969 592 872 96;
  • 13) 0.969 592 872 96 × 2 = 1 + 0.939 185 745 92;
  • 14) 0.939 185 745 92 × 2 = 1 + 0.878 371 491 84;
  • 15) 0.878 371 491 84 × 2 = 1 + 0.756 742 983 68;
  • 16) 0.756 742 983 68 × 2 = 1 + 0.513 485 967 36;
  • 17) 0.513 485 967 36 × 2 = 1 + 0.026 971 934 72;
  • 18) 0.026 971 934 72 × 2 = 0 + 0.053 943 869 44;
  • 19) 0.053 943 869 44 × 2 = 0 + 0.107 887 738 88;
  • 20) 0.107 887 738 88 × 2 = 0 + 0.215 775 477 76;
  • 21) 0.215 775 477 76 × 2 = 0 + 0.431 550 955 52;
  • 22) 0.431 550 955 52 × 2 = 0 + 0.863 101 911 04;
  • 23) 0.863 101 911 04 × 2 = 1 + 0.726 203 822 08;
  • 24) 0.726 203 822 08 × 2 = 1 + 0.452 407 644 16;
  • 25) 0.452 407 644 16 × 2 = 0 + 0.904 815 288 32;
  • 26) 0.904 815 288 32 × 2 = 1 + 0.809 630 576 64;
  • 27) 0.809 630 576 64 × 2 = 1 + 0.619 261 153 28;
  • 28) 0.619 261 153 28 × 2 = 1 + 0.238 522 306 56;
  • 29) 0.238 522 306 56 × 2 = 0 + 0.477 044 613 12;
  • 30) 0.477 044 613 12 × 2 = 0 + 0.954 089 226 24;
  • 31) 0.954 089 226 24 × 2 = 1 + 0.908 178 452 48;
  • 32) 0.908 178 452 48 × 2 = 1 + 0.816 356 904 96;
  • 33) 0.816 356 904 96 × 2 = 1 + 0.632 713 809 92;
  • 34) 0.632 713 809 92 × 2 = 1 + 0.265 427 619 84;
  • 35) 0.265 427 619 84 × 2 = 0 + 0.530 855 239 68;
  • 36) 0.530 855 239 68 × 2 = 1 + 0.061 710 479 36;
  • 37) 0.061 710 479 36 × 2 = 0 + 0.123 420 958 72;
  • 38) 0.123 420 958 72 × 2 = 0 + 0.246 841 917 44;
  • 39) 0.246 841 917 44 × 2 = 0 + 0.493 683 834 88;
  • 40) 0.493 683 834 88 × 2 = 0 + 0.987 367 669 76;
  • 41) 0.987 367 669 76 × 2 = 1 + 0.974 735 339 52;
  • 42) 0.974 735 339 52 × 2 = 1 + 0.949 470 679 04;
  • 43) 0.949 470 679 04 × 2 = 1 + 0.898 941 358 08;
  • 44) 0.898 941 358 08 × 2 = 1 + 0.797 882 716 16;
  • 45) 0.797 882 716 16 × 2 = 1 + 0.595 765 432 32;
  • 46) 0.595 765 432 32 × 2 = 1 + 0.191 530 864 64;
  • 47) 0.191 530 864 64 × 2 = 0 + 0.383 061 729 28;
  • 48) 0.383 061 729 28 × 2 = 0 + 0.766 123 458 56;
  • 49) 0.766 123 458 56 × 2 = 1 + 0.532 246 917 12;
  • 50) 0.532 246 917 12 × 2 = 1 + 0.064 493 834 24;
  • 51) 0.064 493 834 24 × 2 = 0 + 0.128 987 668 48;
  • 52) 0.128 987 668 48 × 2 = 0 + 0.257 975 336 96;
  • 53) 0.257 975 336 96 × 2 = 0 + 0.515 950 673 92;
  • 54) 0.515 950 673 92 × 2 = 1 + 0.031 901 347 84;
  • 55) 0.031 901 347 84 × 2 = 0 + 0.063 802 695 68;
  • 56) 0.063 802 695 68 × 2 = 0 + 0.127 605 391 36;

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 26(10) =

0.0001 1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100(2)

5. Positive number before normalization:

0.110 099 998 26(10) =

0.0001 1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100(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 26(10) =

0.0001 1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100(2) =

0.0001 1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100(2) × 20 =

1.1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100(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 0011 1101 0000 1111 1100 1100 0100


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 0011 1101 0000 1111 1100 1100 0100 =

1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100


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 0011 1101 0000 1111 1100 1100 0100


Decimal number 0.110 099 998 26 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1100 0010 1111 1000 0011 0111 0011 1101 0000 1111 1100 1100 0100

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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