64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 131 094 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 131 094(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. 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;
  • 131 094 ÷ 2 = 65 547 + 0;
  • 65 547 ÷ 2 = 32 773 + 1;
  • 32 773 ÷ 2 = 16 386 + 1;
  • 16 386 ÷ 2 = 8 193 + 0;
  • 8 193 ÷ 2 = 4 096 + 1;
  • 4 096 ÷ 2 = 2 048 + 0;
  • 2 048 ÷ 2 = 1 024 + 0;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 256 ÷ 2 = 128 + 0;
  • 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;

2. Construct the base 2 representation of the positive number.

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


131 094(10) =


10 0000 0000 0001 0110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the left, so that only one non zero digit remains to the left of it:


131 094(10) =


10 0000 0000 0001 0110(2) =


10 0000 0000 0001 0110(2) × 20 =


1.0000 0000 0000 1011 0(2) × 217


4. 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): 17


Mantissa (not normalized):
1.0000 0000 0000 1011 0


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


17 + 2(11-1) - 1 =


(17 + 1 023)(10) =


1 040(10)


6. 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;
  • 1 040 ÷ 2 = 520 + 0;
  • 520 ÷ 2 = 260 + 0;
  • 260 ÷ 2 = 130 + 0;
  • 130 ÷ 2 = 65 + 0;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1040(10) =


100 0001 0000(2)


8. 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 adding the necessary number of zeros to the right.


Mantissa (normalized) =


1. 0 0000 0000 0001 0110 000 0000 0000 0000 0000 0000 0000 0000 0000 =


0000 0000 0000 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000


9. 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) =
100 0001 0000


Mantissa (52 bits) =
0000 0000 0000 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000


The base ten decimal number 131 094 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 0000 - 0000 0000 0000 1011 0000 0000 0000 0000 0000 0000 0000 0000 0000

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