64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 121 212 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 121 212(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;
  • 121 212 ÷ 2 = 60 606 + 0;
  • 60 606 ÷ 2 = 30 303 + 0;
  • 30 303 ÷ 2 = 15 151 + 1;
  • 15 151 ÷ 2 = 7 575 + 1;
  • 7 575 ÷ 2 = 3 787 + 1;
  • 3 787 ÷ 2 = 1 893 + 1;
  • 1 893 ÷ 2 = 946 + 1;
  • 946 ÷ 2 = 473 + 0;
  • 473 ÷ 2 = 236 + 1;
  • 236 ÷ 2 = 118 + 0;
  • 118 ÷ 2 = 59 + 0;
  • 59 ÷ 2 = 29 + 1;
  • 29 ÷ 2 = 14 + 1;
  • 14 ÷ 2 = 7 + 0;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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.


121 212(10) =


1 1101 1001 0111 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the left, so that only one non zero digit remains to the left of it:


121 212(10) =


1 1101 1001 0111 1100(2) =


1 1101 1001 0111 1100(2) × 20 =


1.1101 1001 0111 1100(2) × 216


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): 16


Mantissa (not normalized):
1.1101 1001 0111 1100


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


16 + 2(11-1) - 1 =


(16 + 1 023)(10) =


1 039(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;
  • 1 039 ÷ 2 = 519 + 1;
  • 519 ÷ 2 = 259 + 1;
  • 259 ÷ 2 = 129 + 1;
  • 129 ÷ 2 = 64 + 1;
  • 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;

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) =


1039(10) =


100 0000 1111(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. 1101 1001 0111 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000 =


1101 1001 0111 1100 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 0000 1111


Mantissa (52 bits) =
1101 1001 0111 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000


The base ten decimal number 121 212 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1111 - 1101 1001 0111 1100 0000 0000 0000 0000 0000 0000 0000 0000 0000

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