64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -0.055 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -0.055(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:

|-0.055| = 0.055

2. 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;

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.


0(10) =


0(2)


4. Convert to binary (base 2) the fractional part: 0.055.

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) 0.055 × 2 = 0 + 0.11;
  • 2) 0.11 × 2 = 0 + 0.22;
  • 3) 0.22 × 2 = 0 + 0.44;
  • 4) 0.44 × 2 = 0 + 0.88;
  • 5) 0.88 × 2 = 1 + 0.76;
  • 6) 0.76 × 2 = 1 + 0.52;
  • 7) 0.52 × 2 = 1 + 0.04;
  • 8) 0.04 × 2 = 0 + 0.08;
  • 9) 0.08 × 2 = 0 + 0.16;
  • 10) 0.16 × 2 = 0 + 0.32;
  • 11) 0.32 × 2 = 0 + 0.64;
  • 12) 0.64 × 2 = 1 + 0.28;
  • 13) 0.28 × 2 = 0 + 0.56;
  • 14) 0.56 × 2 = 1 + 0.12;
  • 15) 0.12 × 2 = 0 + 0.24;
  • 16) 0.24 × 2 = 0 + 0.48;
  • 17) 0.48 × 2 = 0 + 0.96;
  • 18) 0.96 × 2 = 1 + 0.92;
  • 19) 0.92 × 2 = 1 + 0.84;
  • 20) 0.84 × 2 = 1 + 0.68;
  • 21) 0.68 × 2 = 1 + 0.36;
  • 22) 0.36 × 2 = 0 + 0.72;
  • 23) 0.72 × 2 = 1 + 0.44;
  • 24) 0.44 × 2 = 0 + 0.88;
  • 25) 0.88 × 2 = 1 + 0.76;
  • 26) 0.76 × 2 = 1 + 0.52;
  • 27) 0.52 × 2 = 1 + 0.04;
  • 28) 0.04 × 2 = 0 + 0.08;
  • 29) 0.08 × 2 = 0 + 0.16;
  • 30) 0.16 × 2 = 0 + 0.32;
  • 31) 0.32 × 2 = 0 + 0.64;
  • 32) 0.64 × 2 = 1 + 0.28;
  • 33) 0.28 × 2 = 0 + 0.56;
  • 34) 0.56 × 2 = 1 + 0.12;
  • 35) 0.12 × 2 = 0 + 0.24;
  • 36) 0.24 × 2 = 0 + 0.48;
  • 37) 0.48 × 2 = 0 + 0.96;
  • 38) 0.96 × 2 = 1 + 0.92;
  • 39) 0.92 × 2 = 1 + 0.84;
  • 40) 0.84 × 2 = 1 + 0.68;
  • 41) 0.68 × 2 = 1 + 0.36;
  • 42) 0.36 × 2 = 0 + 0.72;
  • 43) 0.72 × 2 = 1 + 0.44;
  • 44) 0.44 × 2 = 0 + 0.88;
  • 45) 0.88 × 2 = 1 + 0.76;
  • 46) 0.76 × 2 = 1 + 0.52;
  • 47) 0.52 × 2 = 1 + 0.04;
  • 48) 0.04 × 2 = 0 + 0.08;
  • 49) 0.08 × 2 = 0 + 0.16;
  • 50) 0.16 × 2 = 0 + 0.32;
  • 51) 0.32 × 2 = 0 + 0.64;
  • 52) 0.64 × 2 = 1 + 0.28;
  • 53) 0.28 × 2 = 0 + 0.56;
  • 54) 0.56 × 2 = 1 + 0.12;
  • 55) 0.12 × 2 = 0 + 0.24;
  • 56) 0.24 × 2 = 0 + 0.48;
  • 57) 0.48 × 2 = 0 + 0.96;

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:


0.055(10) =


0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0(2)


6. Positive number before normalization:

0.055(10) =


0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the right, so that only one non zero digit remains to the left of it:


0.055(10) =


0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0(2) =


0.0000 1110 0001 0100 0111 1010 1110 0001 0100 0111 1010 1110 0001 0100 0(2) × 20 =


1.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000(2) × 2-5


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): -5


Mantissa (not normalized):
1.1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


-5 + 2(11-1) - 1 =


(-5 + 1 023)(10) =


1 018(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;
  • 1 018 ÷ 2 = 509 + 0;
  • 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;

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) =


1018(10) =


011 1111 1010(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, only if necessary (not the case here).


Mantissa (normalized) =


1. 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000 =


1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000


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) =
011 1111 1010


Mantissa (52 bits) =
1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000


The base ten decimal number -0.055 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1010 - 1100 0010 1000 1111 0101 1100 0010 1000 1111 0101 1100 0010 1000

The latest decimal numbers converted from base ten to 64 bit double precision IEEE 754 floating point binary standard representation