0.001 219 08 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.001 219 08(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
0.001 219 08(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: 0.001 219 08.

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) 0.001 219 08 × 2 = 0 + 0.002 438 16;
  • 2) 0.002 438 16 × 2 = 0 + 0.004 876 32;
  • 3) 0.004 876 32 × 2 = 0 + 0.009 752 64;
  • 4) 0.009 752 64 × 2 = 0 + 0.019 505 28;
  • 5) 0.019 505 28 × 2 = 0 + 0.039 010 56;
  • 6) 0.039 010 56 × 2 = 0 + 0.078 021 12;
  • 7) 0.078 021 12 × 2 = 0 + 0.156 042 24;
  • 8) 0.156 042 24 × 2 = 0 + 0.312 084 48;
  • 9) 0.312 084 48 × 2 = 0 + 0.624 168 96;
  • 10) 0.624 168 96 × 2 = 1 + 0.248 337 92;
  • 11) 0.248 337 92 × 2 = 0 + 0.496 675 84;
  • 12) 0.496 675 84 × 2 = 0 + 0.993 351 68;
  • 13) 0.993 351 68 × 2 = 1 + 0.986 703 36;
  • 14) 0.986 703 36 × 2 = 1 + 0.973 406 72;
  • 15) 0.973 406 72 × 2 = 1 + 0.946 813 44;
  • 16) 0.946 813 44 × 2 = 1 + 0.893 626 88;
  • 17) 0.893 626 88 × 2 = 1 + 0.787 253 76;
  • 18) 0.787 253 76 × 2 = 1 + 0.574 507 52;
  • 19) 0.574 507 52 × 2 = 1 + 0.149 015 04;
  • 20) 0.149 015 04 × 2 = 0 + 0.298 030 08;
  • 21) 0.298 030 08 × 2 = 0 + 0.596 060 16;
  • 22) 0.596 060 16 × 2 = 1 + 0.192 120 32;
  • 23) 0.192 120 32 × 2 = 0 + 0.384 240 64;
  • 24) 0.384 240 64 × 2 = 0 + 0.768 481 28;
  • 25) 0.768 481 28 × 2 = 1 + 0.536 962 56;
  • 26) 0.536 962 56 × 2 = 1 + 0.073 925 12;
  • 27) 0.073 925 12 × 2 = 0 + 0.147 850 24;
  • 28) 0.147 850 24 × 2 = 0 + 0.295 700 48;
  • 29) 0.295 700 48 × 2 = 0 + 0.591 400 96;
  • 30) 0.591 400 96 × 2 = 1 + 0.182 801 92;
  • 31) 0.182 801 92 × 2 = 0 + 0.365 603 84;
  • 32) 0.365 603 84 × 2 = 0 + 0.731 207 68;
  • 33) 0.731 207 68 × 2 = 1 + 0.462 415 36;
  • 34) 0.462 415 36 × 2 = 0 + 0.924 830 72;
  • 35) 0.924 830 72 × 2 = 1 + 0.849 661 44;
  • 36) 0.849 661 44 × 2 = 1 + 0.699 322 88;
  • 37) 0.699 322 88 × 2 = 1 + 0.398 645 76;
  • 38) 0.398 645 76 × 2 = 0 + 0.797 291 52;
  • 39) 0.797 291 52 × 2 = 1 + 0.594 583 04;
  • 40) 0.594 583 04 × 2 = 1 + 0.189 166 08;
  • 41) 0.189 166 08 × 2 = 0 + 0.378 332 16;
  • 42) 0.378 332 16 × 2 = 0 + 0.756 664 32;
  • 43) 0.756 664 32 × 2 = 1 + 0.513 328 64;
  • 44) 0.513 328 64 × 2 = 1 + 0.026 657 28;
  • 45) 0.026 657 28 × 2 = 0 + 0.053 314 56;
  • 46) 0.053 314 56 × 2 = 0 + 0.106 629 12;
  • 47) 0.106 629 12 × 2 = 0 + 0.213 258 24;
  • 48) 0.213 258 24 × 2 = 0 + 0.426 516 48;
  • 49) 0.426 516 48 × 2 = 0 + 0.853 032 96;
  • 50) 0.853 032 96 × 2 = 1 + 0.706 065 92;
  • 51) 0.706 065 92 × 2 = 1 + 0.412 131 84;
  • 52) 0.412 131 84 × 2 = 0 + 0.824 263 68;
  • 53) 0.824 263 68 × 2 = 1 + 0.648 527 36;
  • 54) 0.648 527 36 × 2 = 1 + 0.297 054 72;
  • 55) 0.297 054 72 × 2 = 0 + 0.594 109 44;
  • 56) 0.594 109 44 × 2 = 1 + 0.188 218 88;
  • 57) 0.188 218 88 × 2 = 0 + 0.376 437 76;
  • 58) 0.376 437 76 × 2 = 0 + 0.752 875 52;
  • 59) 0.752 875 52 × 2 = 1 + 0.505 751 04;
  • 60) 0.505 751 04 × 2 = 1 + 0.011 502 08;
  • 61) 0.011 502 08 × 2 = 0 + 0.023 004 16;
  • 62) 0.023 004 16 × 2 = 0 + 0.046 008 32;

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:


0.001 219 08(10) =

0.0000 0000 0100 1111 1110 0100 1100 0100 1011 1011 0011 0000 0110 1101 0011 00(2)

5. Positive number before normalization:

0.001 219 08(10) =

0.0000 0000 0100 1111 1110 0100 1100 0100 1011 1011 0011 0000 0110 1101 0011 00(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the right, so that only one non zero digit remains to the left of it:


0.001 219 08(10) =

0.0000 0000 0100 1111 1110 0100 1100 0100 1011 1011 0011 0000 0110 1101 0011 00(2) =

0.0000 0000 0100 1111 1110 0100 1100 0100 1011 1011 0011 0000 0110 1101 0011 00(2) × 20 =

1.0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100(2) × 2-10


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): -10

Mantissa (not normalized):
1.0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-10 + 2(11-1) - 1 =

(-10 + 1 023)(10) =

1 013(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;
  • 1 013 ÷ 2 = 506 + 1;
  • 506 ÷ 2 = 253 + 0;
  • 253 ÷ 2 = 126 + 1;
  • 126 ÷ 2 = 63 + 0;
  • 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) =

1013(10) =

011 1111 0101(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. 0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100 =

0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100


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 0101

Mantissa (52 bits) =
0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100


Decimal number 0.001 219 08 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 0101 - 0011 1111 1001 0011 0001 0010 1110 1100 1100 0001 1011 0100 1100

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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