64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -128.857 36 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -128.857 36(10) converted and written in 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. Start with the positive version of the number:

|-128.857 36| = 128.857 36

2. First, convert to binary (in base 2) the integer part: 128.
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;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.


128(10) =


1000 0000(2)


4. Convert to binary (base 2) the fractional part: 0.857 36.

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) 0.857 36 × 2 = 1 + 0.714 72;
  • 2) 0.714 72 × 2 = 1 + 0.429 44;
  • 3) 0.429 44 × 2 = 0 + 0.858 88;
  • 4) 0.858 88 × 2 = 1 + 0.717 76;
  • 5) 0.717 76 × 2 = 1 + 0.435 52;
  • 6) 0.435 52 × 2 = 0 + 0.871 04;
  • 7) 0.871 04 × 2 = 1 + 0.742 08;
  • 8) 0.742 08 × 2 = 1 + 0.484 16;
  • 9) 0.484 16 × 2 = 0 + 0.968 32;
  • 10) 0.968 32 × 2 = 1 + 0.936 64;
  • 11) 0.936 64 × 2 = 1 + 0.873 28;
  • 12) 0.873 28 × 2 = 1 + 0.746 56;
  • 13) 0.746 56 × 2 = 1 + 0.493 12;
  • 14) 0.493 12 × 2 = 0 + 0.986 24;
  • 15) 0.986 24 × 2 = 1 + 0.972 48;
  • 16) 0.972 48 × 2 = 1 + 0.944 96;
  • 17) 0.944 96 × 2 = 1 + 0.889 92;
  • 18) 0.889 92 × 2 = 1 + 0.779 84;
  • 19) 0.779 84 × 2 = 1 + 0.559 68;
  • 20) 0.559 68 × 2 = 1 + 0.119 36;
  • 21) 0.119 36 × 2 = 0 + 0.238 72;
  • 22) 0.238 72 × 2 = 0 + 0.477 44;
  • 23) 0.477 44 × 2 = 0 + 0.954 88;
  • 24) 0.954 88 × 2 = 1 + 0.909 76;
  • 25) 0.909 76 × 2 = 1 + 0.819 52;
  • 26) 0.819 52 × 2 = 1 + 0.639 04;
  • 27) 0.639 04 × 2 = 1 + 0.278 08;
  • 28) 0.278 08 × 2 = 0 + 0.556 16;
  • 29) 0.556 16 × 2 = 1 + 0.112 32;
  • 30) 0.112 32 × 2 = 0 + 0.224 64;
  • 31) 0.224 64 × 2 = 0 + 0.449 28;
  • 32) 0.449 28 × 2 = 0 + 0.898 56;
  • 33) 0.898 56 × 2 = 1 + 0.797 12;
  • 34) 0.797 12 × 2 = 1 + 0.594 24;
  • 35) 0.594 24 × 2 = 1 + 0.188 48;
  • 36) 0.188 48 × 2 = 0 + 0.376 96;
  • 37) 0.376 96 × 2 = 0 + 0.753 92;
  • 38) 0.753 92 × 2 = 1 + 0.507 84;
  • 39) 0.507 84 × 2 = 1 + 0.015 68;
  • 40) 0.015 68 × 2 = 0 + 0.031 36;
  • 41) 0.031 36 × 2 = 0 + 0.062 72;
  • 42) 0.062 72 × 2 = 0 + 0.125 44;
  • 43) 0.125 44 × 2 = 0 + 0.250 88;
  • 44) 0.250 88 × 2 = 0 + 0.501 76;
  • 45) 0.501 76 × 2 = 1 + 0.003 52;
  • 46) 0.003 52 × 2 = 0 + 0.007 04;
  • 47) 0.007 04 × 2 = 0 + 0.014 08;
  • 48) 0.014 08 × 2 = 0 + 0.028 16;
  • 49) 0.028 16 × 2 = 0 + 0.056 32;
  • 50) 0.056 32 × 2 = 0 + 0.112 64;
  • 51) 0.112 64 × 2 = 0 + 0.225 28;
  • 52) 0.225 28 × 2 = 0 + 0.450 56;
  • 53) 0.450 56 × 2 = 0 + 0.901 12;

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...)


5. 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:


0.857 36(10) =


0.1101 1011 0111 1011 1111 0001 1110 1000 1110 0110 0000 1000 0000 0(2)


6. Positive number before normalization:

128.857 36(10) =


1000 0000.1101 1011 0111 1011 1111 0001 1110 1000 1110 0110 0000 1000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


128.857 36(10) =


1000 0000.1101 1011 0111 1011 1111 0001 1110 1000 1110 0110 0000 1000 0000 0(2) =


1000 0000.1101 1011 0111 1011 1111 0001 1110 1000 1110 0110 0000 1000 0000 0(2) × 20 =


1.0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001 0000 0000(2) × 27


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): 7


Mantissa (not normalized):
1.0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001 0000 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


7 + 2(11-1) - 1 =


(7 + 1 023)(10) =


1 030(10)


10. 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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 128 ÷ 2 = 64 + 0;
  • 64 ÷ 2 = 32 + 0;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

11. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =


1030(10) =


100 0000 0110(2)


12. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001 0000 0000 =


0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
100 0000 0110


Mantissa (52 bits) =
0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001


The base ten decimal number -128.857 36 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0110 - 0000 0001 1011 0110 1111 0111 1110 0011 1101 0001 1100 1100 0001

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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