64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 842 019 118 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 842 019 118(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;
  • 842 019 118 ÷ 2 = 421 009 559 + 0;
  • 421 009 559 ÷ 2 = 210 504 779 + 1;
  • 210 504 779 ÷ 2 = 105 252 389 + 1;
  • 105 252 389 ÷ 2 = 52 626 194 + 1;
  • 52 626 194 ÷ 2 = 26 313 097 + 0;
  • 26 313 097 ÷ 2 = 13 156 548 + 1;
  • 13 156 548 ÷ 2 = 6 578 274 + 0;
  • 6 578 274 ÷ 2 = 3 289 137 + 0;
  • 3 289 137 ÷ 2 = 1 644 568 + 1;
  • 1 644 568 ÷ 2 = 822 284 + 0;
  • 822 284 ÷ 2 = 411 142 + 0;
  • 411 142 ÷ 2 = 205 571 + 0;
  • 205 571 ÷ 2 = 102 785 + 1;
  • 102 785 ÷ 2 = 51 392 + 1;
  • 51 392 ÷ 2 = 25 696 + 0;
  • 25 696 ÷ 2 = 12 848 + 0;
  • 12 848 ÷ 2 = 6 424 + 0;
  • 6 424 ÷ 2 = 3 212 + 0;
  • 3 212 ÷ 2 = 1 606 + 0;
  • 1 606 ÷ 2 = 803 + 0;
  • 803 ÷ 2 = 401 + 1;
  • 401 ÷ 2 = 200 + 1;
  • 200 ÷ 2 = 100 + 0;
  • 100 ÷ 2 = 50 + 0;
  • 50 ÷ 2 = 25 + 0;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 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.


842 019 118(10) =


11 0010 0011 0000 0011 0001 0010 1110(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 29 positions to the left, so that only one non zero digit remains to the left of it:


842 019 118(10) =


11 0010 0011 0000 0011 0001 0010 1110(2) =


11 0010 0011 0000 0011 0001 0010 1110(2) × 20 =


1.1001 0001 1000 0001 1000 1001 0111 0(2) × 229


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): 29


Mantissa (not normalized):
1.1001 0001 1000 0001 1000 1001 0111 0


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


29 + 2(11-1) - 1 =


(29 + 1 023)(10) =


1 052(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;
  • 1 052 ÷ 2 = 526 + 0;
  • 526 ÷ 2 = 263 + 0;
  • 263 ÷ 2 = 131 + 1;
  • 131 ÷ 2 = 65 + 1;
  • 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) =


1052(10) =


100 0001 1100(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. 1 0010 0011 0000 0011 0001 0010 1110 000 0000 0000 0000 0000 0000 =


1001 0001 1000 0001 1000 1001 0111 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 1100


Mantissa (52 bits) =
1001 0001 1000 0001 1000 1001 0111 0000 0000 0000 0000 0000 0000


The base ten decimal number 842 019 118 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1100 - 1001 0001 1000 0001 1000 1001 0111 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