0.119 999 999 999 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.119 999 999 999 9(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
0.119 999 999 999 9(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: 0.119 999 999 999 9.

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) 0.119 999 999 999 9 × 2 = 0 + 0.239 999 999 999 8;
  • 2) 0.239 999 999 999 8 × 2 = 0 + 0.479 999 999 999 6;
  • 3) 0.479 999 999 999 6 × 2 = 0 + 0.959 999 999 999 2;
  • 4) 0.959 999 999 999 2 × 2 = 1 + 0.919 999 999 998 4;
  • 5) 0.919 999 999 998 4 × 2 = 1 + 0.839 999 999 996 8;
  • 6) 0.839 999 999 996 8 × 2 = 1 + 0.679 999 999 993 6;
  • 7) 0.679 999 999 993 6 × 2 = 1 + 0.359 999 999 987 2;
  • 8) 0.359 999 999 987 2 × 2 = 0 + 0.719 999 999 974 4;
  • 9) 0.719 999 999 974 4 × 2 = 1 + 0.439 999 999 948 8;
  • 10) 0.439 999 999 948 8 × 2 = 0 + 0.879 999 999 897 6;
  • 11) 0.879 999 999 897 6 × 2 = 1 + 0.759 999 999 795 2;
  • 12) 0.759 999 999 795 2 × 2 = 1 + 0.519 999 999 590 4;
  • 13) 0.519 999 999 590 4 × 2 = 1 + 0.039 999 999 180 8;
  • 14) 0.039 999 999 180 8 × 2 = 0 + 0.079 999 998 361 6;
  • 15) 0.079 999 998 361 6 × 2 = 0 + 0.159 999 996 723 2;
  • 16) 0.159 999 996 723 2 × 2 = 0 + 0.319 999 993 446 4;
  • 17) 0.319 999 993 446 4 × 2 = 0 + 0.639 999 986 892 8;
  • 18) 0.639 999 986 892 8 × 2 = 1 + 0.279 999 973 785 6;
  • 19) 0.279 999 973 785 6 × 2 = 0 + 0.559 999 947 571 2;
  • 20) 0.559 999 947 571 2 × 2 = 1 + 0.119 999 895 142 4;
  • 21) 0.119 999 895 142 4 × 2 = 0 + 0.239 999 790 284 8;
  • 22) 0.239 999 790 284 8 × 2 = 0 + 0.479 999 580 569 6;
  • 23) 0.479 999 580 569 6 × 2 = 0 + 0.959 999 161 139 2;
  • 24) 0.959 999 161 139 2 × 2 = 1 + 0.919 998 322 278 4;
  • 25) 0.919 998 322 278 4 × 2 = 1 + 0.839 996 644 556 8;
  • 26) 0.839 996 644 556 8 × 2 = 1 + 0.679 993 289 113 6;
  • 27) 0.679 993 289 113 6 × 2 = 1 + 0.359 986 578 227 2;
  • 28) 0.359 986 578 227 2 × 2 = 0 + 0.719 973 156 454 4;
  • 29) 0.719 973 156 454 4 × 2 = 1 + 0.439 946 312 908 8;
  • 30) 0.439 946 312 908 8 × 2 = 0 + 0.879 892 625 817 6;
  • 31) 0.879 892 625 817 6 × 2 = 1 + 0.759 785 251 635 2;
  • 32) 0.759 785 251 635 2 × 2 = 1 + 0.519 570 503 270 4;
  • 33) 0.519 570 503 270 4 × 2 = 1 + 0.039 141 006 540 8;
  • 34) 0.039 141 006 540 8 × 2 = 0 + 0.078 282 013 081 6;
  • 35) 0.078 282 013 081 6 × 2 = 0 + 0.156 564 026 163 2;
  • 36) 0.156 564 026 163 2 × 2 = 0 + 0.313 128 052 326 4;
  • 37) 0.313 128 052 326 4 × 2 = 0 + 0.626 256 104 652 8;
  • 38) 0.626 256 104 652 8 × 2 = 1 + 0.252 512 209 305 6;
  • 39) 0.252 512 209 305 6 × 2 = 0 + 0.505 024 418 611 2;
  • 40) 0.505 024 418 611 2 × 2 = 1 + 0.010 048 837 222 4;
  • 41) 0.010 048 837 222 4 × 2 = 0 + 0.020 097 674 444 8;
  • 42) 0.020 097 674 444 8 × 2 = 0 + 0.040 195 348 889 6;
  • 43) 0.040 195 348 889 6 × 2 = 0 + 0.080 390 697 779 2;
  • 44) 0.080 390 697 779 2 × 2 = 0 + 0.160 781 395 558 4;
  • 45) 0.160 781 395 558 4 × 2 = 0 + 0.321 562 791 116 8;
  • 46) 0.321 562 791 116 8 × 2 = 0 + 0.643 125 582 233 6;
  • 47) 0.643 125 582 233 6 × 2 = 1 + 0.286 251 164 467 2;
  • 48) 0.286 251 164 467 2 × 2 = 0 + 0.572 502 328 934 4;
  • 49) 0.572 502 328 934 4 × 2 = 1 + 0.145 004 657 868 8;
  • 50) 0.145 004 657 868 8 × 2 = 0 + 0.290 009 315 737 6;
  • 51) 0.290 009 315 737 6 × 2 = 0 + 0.580 018 631 475 2;
  • 52) 0.580 018 631 475 2 × 2 = 1 + 0.160 037 262 950 4;
  • 53) 0.160 037 262 950 4 × 2 = 0 + 0.320 074 525 900 8;
  • 54) 0.320 074 525 900 8 × 2 = 0 + 0.640 149 051 801 6;
  • 55) 0.640 149 051 801 6 × 2 = 1 + 0.280 298 103 603 2;
  • 56) 0.280 298 103 603 2 × 2 = 0 + 0.560 596 207 206 4;

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:


0.119 999 999 999 9(10) =

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

5. Positive number before normalization:

0.119 999 999 999 9(10) =

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


0.119 999 999 999 9(10) =

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

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

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


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 0101 0000 0010 1001 0010 =

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


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 0101 0000 0010 1001 0010


Decimal number 0.119 999 999 999 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

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

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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