0.110 099 999 32 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.110 099 999 32(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
0.110 099 999 32(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: 0.110 099 999 32.

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) 0.110 099 999 32 × 2 = 0 + 0.220 199 998 64;
  • 2) 0.220 199 998 64 × 2 = 0 + 0.440 399 997 28;
  • 3) 0.440 399 997 28 × 2 = 0 + 0.880 799 994 56;
  • 4) 0.880 799 994 56 × 2 = 1 + 0.761 599 989 12;
  • 5) 0.761 599 989 12 × 2 = 1 + 0.523 199 978 24;
  • 6) 0.523 199 978 24 × 2 = 1 + 0.046 399 956 48;
  • 7) 0.046 399 956 48 × 2 = 0 + 0.092 799 912 96;
  • 8) 0.092 799 912 96 × 2 = 0 + 0.185 599 825 92;
  • 9) 0.185 599 825 92 × 2 = 0 + 0.371 199 651 84;
  • 10) 0.371 199 651 84 × 2 = 0 + 0.742 399 303 68;
  • 11) 0.742 399 303 68 × 2 = 1 + 0.484 798 607 36;
  • 12) 0.484 798 607 36 × 2 = 0 + 0.969 597 214 72;
  • 13) 0.969 597 214 72 × 2 = 1 + 0.939 194 429 44;
  • 14) 0.939 194 429 44 × 2 = 1 + 0.878 388 858 88;
  • 15) 0.878 388 858 88 × 2 = 1 + 0.756 777 717 76;
  • 16) 0.756 777 717 76 × 2 = 1 + 0.513 555 435 52;
  • 17) 0.513 555 435 52 × 2 = 1 + 0.027 110 871 04;
  • 18) 0.027 110 871 04 × 2 = 0 + 0.054 221 742 08;
  • 19) 0.054 221 742 08 × 2 = 0 + 0.108 443 484 16;
  • 20) 0.108 443 484 16 × 2 = 0 + 0.216 886 968 32;
  • 21) 0.216 886 968 32 × 2 = 0 + 0.433 773 936 64;
  • 22) 0.433 773 936 64 × 2 = 0 + 0.867 547 873 28;
  • 23) 0.867 547 873 28 × 2 = 1 + 0.735 095 746 56;
  • 24) 0.735 095 746 56 × 2 = 1 + 0.470 191 493 12;
  • 25) 0.470 191 493 12 × 2 = 0 + 0.940 382 986 24;
  • 26) 0.940 382 986 24 × 2 = 1 + 0.880 765 972 48;
  • 27) 0.880 765 972 48 × 2 = 1 + 0.761 531 944 96;
  • 28) 0.761 531 944 96 × 2 = 1 + 0.523 063 889 92;
  • 29) 0.523 063 889 92 × 2 = 1 + 0.046 127 779 84;
  • 30) 0.046 127 779 84 × 2 = 0 + 0.092 255 559 68;
  • 31) 0.092 255 559 68 × 2 = 0 + 0.184 511 119 36;
  • 32) 0.184 511 119 36 × 2 = 0 + 0.369 022 238 72;
  • 33) 0.369 022 238 72 × 2 = 0 + 0.738 044 477 44;
  • 34) 0.738 044 477 44 × 2 = 1 + 0.476 088 954 88;
  • 35) 0.476 088 954 88 × 2 = 0 + 0.952 177 909 76;
  • 36) 0.952 177 909 76 × 2 = 1 + 0.904 355 819 52;
  • 37) 0.904 355 819 52 × 2 = 1 + 0.808 711 639 04;
  • 38) 0.808 711 639 04 × 2 = 1 + 0.617 423 278 08;
  • 39) 0.617 423 278 08 × 2 = 1 + 0.234 846 556 16;
  • 40) 0.234 846 556 16 × 2 = 0 + 0.469 693 112 32;
  • 41) 0.469 693 112 32 × 2 = 0 + 0.939 386 224 64;
  • 42) 0.939 386 224 64 × 2 = 1 + 0.878 772 449 28;
  • 43) 0.878 772 449 28 × 2 = 1 + 0.757 544 898 56;
  • 44) 0.757 544 898 56 × 2 = 1 + 0.515 089 797 12;
  • 45) 0.515 089 797 12 × 2 = 1 + 0.030 179 594 24;
  • 46) 0.030 179 594 24 × 2 = 0 + 0.060 359 188 48;
  • 47) 0.060 359 188 48 × 2 = 0 + 0.120 718 376 96;
  • 48) 0.120 718 376 96 × 2 = 0 + 0.241 436 753 92;
  • 49) 0.241 436 753 92 × 2 = 0 + 0.482 873 507 84;
  • 50) 0.482 873 507 84 × 2 = 0 + 0.965 747 015 68;
  • 51) 0.965 747 015 68 × 2 = 1 + 0.931 494 031 36;
  • 52) 0.931 494 031 36 × 2 = 1 + 0.862 988 062 72;
  • 53) 0.862 988 062 72 × 2 = 1 + 0.725 976 125 44;
  • 54) 0.725 976 125 44 × 2 = 1 + 0.451 952 250 88;
  • 55) 0.451 952 250 88 × 2 = 0 + 0.903 904 501 76;
  • 56) 0.903 904 501 76 × 2 = 1 + 0.807 809 003 52;

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:


0.110 099 999 32(10) =

0.0001 1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101(2)

5. Positive number before normalization:

0.110 099 999 32(10) =

0.0001 1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101(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:


0.110 099 999 32(10) =

0.0001 1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101(2) =

0.0001 1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101(2) × 20 =

1.1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101(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 1000 0101 1110 0111 1000 0011 1101


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 1000 0101 1110 0111 1000 0011 1101 =

1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101


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 1000 0101 1110 0111 1000 0011 1101


Decimal number 0.110 099 999 32 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1100 0010 1111 1000 0011 0111 1000 0101 1110 0111 1000 0011 1101

» Convert decimal 0.110 099 999 33 to 64 bit double precision IEEE 754 binary floating point representation standard

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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