0.119 999 999 998 1 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 998 1(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
0.119 999 999 998 1(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: 0.119 999 999 998 1.

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) 0.119 999 999 998 1 × 2 = 0 + 0.239 999 999 996 2;
  • 2) 0.239 999 999 996 2 × 2 = 0 + 0.479 999 999 992 4;
  • 3) 0.479 999 999 992 4 × 2 = 0 + 0.959 999 999 984 8;
  • 4) 0.959 999 999 984 8 × 2 = 1 + 0.919 999 999 969 6;
  • 5) 0.919 999 999 969 6 × 2 = 1 + 0.839 999 999 939 2;
  • 6) 0.839 999 999 939 2 × 2 = 1 + 0.679 999 999 878 4;
  • 7) 0.679 999 999 878 4 × 2 = 1 + 0.359 999 999 756 8;
  • 8) 0.359 999 999 756 8 × 2 = 0 + 0.719 999 999 513 6;
  • 9) 0.719 999 999 513 6 × 2 = 1 + 0.439 999 999 027 2;
  • 10) 0.439 999 999 027 2 × 2 = 0 + 0.879 999 998 054 4;
  • 11) 0.879 999 998 054 4 × 2 = 1 + 0.759 999 996 108 8;
  • 12) 0.759 999 996 108 8 × 2 = 1 + 0.519 999 992 217 6;
  • 13) 0.519 999 992 217 6 × 2 = 1 + 0.039 999 984 435 2;
  • 14) 0.039 999 984 435 2 × 2 = 0 + 0.079 999 968 870 4;
  • 15) 0.079 999 968 870 4 × 2 = 0 + 0.159 999 937 740 8;
  • 16) 0.159 999 937 740 8 × 2 = 0 + 0.319 999 875 481 6;
  • 17) 0.319 999 875 481 6 × 2 = 0 + 0.639 999 750 963 2;
  • 18) 0.639 999 750 963 2 × 2 = 1 + 0.279 999 501 926 4;
  • 19) 0.279 999 501 926 4 × 2 = 0 + 0.559 999 003 852 8;
  • 20) 0.559 999 003 852 8 × 2 = 1 + 0.119 998 007 705 6;
  • 21) 0.119 998 007 705 6 × 2 = 0 + 0.239 996 015 411 2;
  • 22) 0.239 996 015 411 2 × 2 = 0 + 0.479 992 030 822 4;
  • 23) 0.479 992 030 822 4 × 2 = 0 + 0.959 984 061 644 8;
  • 24) 0.959 984 061 644 8 × 2 = 1 + 0.919 968 123 289 6;
  • 25) 0.919 968 123 289 6 × 2 = 1 + 0.839 936 246 579 2;
  • 26) 0.839 936 246 579 2 × 2 = 1 + 0.679 872 493 158 4;
  • 27) 0.679 872 493 158 4 × 2 = 1 + 0.359 744 986 316 8;
  • 28) 0.359 744 986 316 8 × 2 = 0 + 0.719 489 972 633 6;
  • 29) 0.719 489 972 633 6 × 2 = 1 + 0.438 979 945 267 2;
  • 30) 0.438 979 945 267 2 × 2 = 0 + 0.877 959 890 534 4;
  • 31) 0.877 959 890 534 4 × 2 = 1 + 0.755 919 781 068 8;
  • 32) 0.755 919 781 068 8 × 2 = 1 + 0.511 839 562 137 6;
  • 33) 0.511 839 562 137 6 × 2 = 1 + 0.023 679 124 275 2;
  • 34) 0.023 679 124 275 2 × 2 = 0 + 0.047 358 248 550 4;
  • 35) 0.047 358 248 550 4 × 2 = 0 + 0.094 716 497 100 8;
  • 36) 0.094 716 497 100 8 × 2 = 0 + 0.189 432 994 201 6;
  • 37) 0.189 432 994 201 6 × 2 = 0 + 0.378 865 988 403 2;
  • 38) 0.378 865 988 403 2 × 2 = 0 + 0.757 731 976 806 4;
  • 39) 0.757 731 976 806 4 × 2 = 1 + 0.515 463 953 612 8;
  • 40) 0.515 463 953 612 8 × 2 = 1 + 0.030 927 907 225 6;
  • 41) 0.030 927 907 225 6 × 2 = 0 + 0.061 855 814 451 2;
  • 42) 0.061 855 814 451 2 × 2 = 0 + 0.123 711 628 902 4;
  • 43) 0.123 711 628 902 4 × 2 = 0 + 0.247 423 257 804 8;
  • 44) 0.247 423 257 804 8 × 2 = 0 + 0.494 846 515 609 6;
  • 45) 0.494 846 515 609 6 × 2 = 0 + 0.989 693 031 219 2;
  • 46) 0.989 693 031 219 2 × 2 = 1 + 0.979 386 062 438 4;
  • 47) 0.979 386 062 438 4 × 2 = 1 + 0.958 772 124 876 8;
  • 48) 0.958 772 124 876 8 × 2 = 1 + 0.917 544 249 753 6;
  • 49) 0.917 544 249 753 6 × 2 = 1 + 0.835 088 499 507 2;
  • 50) 0.835 088 499 507 2 × 2 = 1 + 0.670 176 999 014 4;
  • 51) 0.670 176 999 014 4 × 2 = 1 + 0.340 353 998 028 8;
  • 52) 0.340 353 998 028 8 × 2 = 0 + 0.680 707 996 057 6;
  • 53) 0.680 707 996 057 6 × 2 = 1 + 0.361 415 992 115 2;
  • 54) 0.361 415 992 115 2 × 2 = 0 + 0.722 831 984 230 4;
  • 55) 0.722 831 984 230 4 × 2 = 1 + 0.445 663 968 460 8;
  • 56) 0.445 663 968 460 8 × 2 = 0 + 0.891 327 936 921 6;

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:


0.119 999 999 998 1(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010(2)

5. Positive number before normalization:

0.119 999 999 998 1(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010(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:


0.119 999 999 998 1(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010(2) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010(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):
1.1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010


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. 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010 =

1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010


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) =
1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010


Decimal number 0.119 999 999 998 1 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 1000 0011 0000 0111 1110 1010

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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