0.119 999 999 997 2 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 997 2(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
0.119 999 999 997 2(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: 0.119 999 999 997 2.

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) 0.119 999 999 997 2 × 2 = 0 + 0.239 999 999 994 4;
  • 2) 0.239 999 999 994 4 × 2 = 0 + 0.479 999 999 988 8;
  • 3) 0.479 999 999 988 8 × 2 = 0 + 0.959 999 999 977 6;
  • 4) 0.959 999 999 977 6 × 2 = 1 + 0.919 999 999 955 2;
  • 5) 0.919 999 999 955 2 × 2 = 1 + 0.839 999 999 910 4;
  • 6) 0.839 999 999 910 4 × 2 = 1 + 0.679 999 999 820 8;
  • 7) 0.679 999 999 820 8 × 2 = 1 + 0.359 999 999 641 6;
  • 8) 0.359 999 999 641 6 × 2 = 0 + 0.719 999 999 283 2;
  • 9) 0.719 999 999 283 2 × 2 = 1 + 0.439 999 998 566 4;
  • 10) 0.439 999 998 566 4 × 2 = 0 + 0.879 999 997 132 8;
  • 11) 0.879 999 997 132 8 × 2 = 1 + 0.759 999 994 265 6;
  • 12) 0.759 999 994 265 6 × 2 = 1 + 0.519 999 988 531 2;
  • 13) 0.519 999 988 531 2 × 2 = 1 + 0.039 999 977 062 4;
  • 14) 0.039 999 977 062 4 × 2 = 0 + 0.079 999 954 124 8;
  • 15) 0.079 999 954 124 8 × 2 = 0 + 0.159 999 908 249 6;
  • 16) 0.159 999 908 249 6 × 2 = 0 + 0.319 999 816 499 2;
  • 17) 0.319 999 816 499 2 × 2 = 0 + 0.639 999 632 998 4;
  • 18) 0.639 999 632 998 4 × 2 = 1 + 0.279 999 265 996 8;
  • 19) 0.279 999 265 996 8 × 2 = 0 + 0.559 998 531 993 6;
  • 20) 0.559 998 531 993 6 × 2 = 1 + 0.119 997 063 987 2;
  • 21) 0.119 997 063 987 2 × 2 = 0 + 0.239 994 127 974 4;
  • 22) 0.239 994 127 974 4 × 2 = 0 + 0.479 988 255 948 8;
  • 23) 0.479 988 255 948 8 × 2 = 0 + 0.959 976 511 897 6;
  • 24) 0.959 976 511 897 6 × 2 = 1 + 0.919 953 023 795 2;
  • 25) 0.919 953 023 795 2 × 2 = 1 + 0.839 906 047 590 4;
  • 26) 0.839 906 047 590 4 × 2 = 1 + 0.679 812 095 180 8;
  • 27) 0.679 812 095 180 8 × 2 = 1 + 0.359 624 190 361 6;
  • 28) 0.359 624 190 361 6 × 2 = 0 + 0.719 248 380 723 2;
  • 29) 0.719 248 380 723 2 × 2 = 1 + 0.438 496 761 446 4;
  • 30) 0.438 496 761 446 4 × 2 = 0 + 0.876 993 522 892 8;
  • 31) 0.876 993 522 892 8 × 2 = 1 + 0.753 987 045 785 6;
  • 32) 0.753 987 045 785 6 × 2 = 1 + 0.507 974 091 571 2;
  • 33) 0.507 974 091 571 2 × 2 = 1 + 0.015 948 183 142 4;
  • 34) 0.015 948 183 142 4 × 2 = 0 + 0.031 896 366 284 8;
  • 35) 0.031 896 366 284 8 × 2 = 0 + 0.063 792 732 569 6;
  • 36) 0.063 792 732 569 6 × 2 = 0 + 0.127 585 465 139 2;
  • 37) 0.127 585 465 139 2 × 2 = 0 + 0.255 170 930 278 4;
  • 38) 0.255 170 930 278 4 × 2 = 0 + 0.510 341 860 556 8;
  • 39) 0.510 341 860 556 8 × 2 = 1 + 0.020 683 721 113 6;
  • 40) 0.020 683 721 113 6 × 2 = 0 + 0.041 367 442 227 2;
  • 41) 0.041 367 442 227 2 × 2 = 0 + 0.082 734 884 454 4;
  • 42) 0.082 734 884 454 4 × 2 = 0 + 0.165 469 768 908 8;
  • 43) 0.165 469 768 908 8 × 2 = 0 + 0.330 939 537 817 6;
  • 44) 0.330 939 537 817 6 × 2 = 0 + 0.661 879 075 635 2;
  • 45) 0.661 879 075 635 2 × 2 = 1 + 0.323 758 151 270 4;
  • 46) 0.323 758 151 270 4 × 2 = 0 + 0.647 516 302 540 8;
  • 47) 0.647 516 302 540 8 × 2 = 1 + 0.295 032 605 081 6;
  • 48) 0.295 032 605 081 6 × 2 = 0 + 0.590 065 210 163 2;
  • 49) 0.590 065 210 163 2 × 2 = 1 + 0.180 130 420 326 4;
  • 50) 0.180 130 420 326 4 × 2 = 0 + 0.360 260 840 652 8;
  • 51) 0.360 260 840 652 8 × 2 = 0 + 0.720 521 681 305 6;
  • 52) 0.720 521 681 305 6 × 2 = 1 + 0.441 043 362 611 2;
  • 53) 0.441 043 362 611 2 × 2 = 0 + 0.882 086 725 222 4;
  • 54) 0.882 086 725 222 4 × 2 = 1 + 0.764 173 450 444 8;
  • 55) 0.764 173 450 444 8 × 2 = 1 + 0.528 346 900 889 6;
  • 56) 0.528 346 900 889 6 × 2 = 1 + 0.056 693 801 779 2;

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:


0.119 999 999 997 2(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111(2)

5. Positive number before normalization:

0.119 999 999 997 2(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111(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:


0.119 999 999 997 2(10) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111(2) =

0.0001 1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111(2) × 20 =

1.1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111(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 0010 0000 1010 1001 0111


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 0010 0000 1010 1001 0111 =

1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111


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 0010 0000 1010 1001 0111


Decimal number 0.119 999 999 997 2 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1011 - 1110 1011 1000 0101 0001 1110 1011 1000 0010 0000 1010 1001 0111

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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