64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 141 982 385 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 141 982 385(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;
  • 141 982 385 ÷ 2 = 70 991 192 + 1;
  • 70 991 192 ÷ 2 = 35 495 596 + 0;
  • 35 495 596 ÷ 2 = 17 747 798 + 0;
  • 17 747 798 ÷ 2 = 8 873 899 + 0;
  • 8 873 899 ÷ 2 = 4 436 949 + 1;
  • 4 436 949 ÷ 2 = 2 218 474 + 1;
  • 2 218 474 ÷ 2 = 1 109 237 + 0;
  • 1 109 237 ÷ 2 = 554 618 + 1;
  • 554 618 ÷ 2 = 277 309 + 0;
  • 277 309 ÷ 2 = 138 654 + 1;
  • 138 654 ÷ 2 = 69 327 + 0;
  • 69 327 ÷ 2 = 34 663 + 1;
  • 34 663 ÷ 2 = 17 331 + 1;
  • 17 331 ÷ 2 = 8 665 + 1;
  • 8 665 ÷ 2 = 4 332 + 1;
  • 4 332 ÷ 2 = 2 166 + 0;
  • 2 166 ÷ 2 = 1 083 + 0;
  • 1 083 ÷ 2 = 541 + 1;
  • 541 ÷ 2 = 270 + 1;
  • 270 ÷ 2 = 135 + 0;
  • 135 ÷ 2 = 67 + 1;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 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.


141 982 385(10) =


1000 0111 0110 0111 1010 1011 0001(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the left, so that only one non zero digit remains to the left of it:


141 982 385(10) =


1000 0111 0110 0111 1010 1011 0001(2) =


1000 0111 0110 0111 1010 1011 0001(2) × 20 =


1.0000 1110 1100 1111 0101 0110 001(2) × 227


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): 27


Mantissa (not normalized):
1.0000 1110 1100 1111 0101 0110 001


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


27 + 2(11-1) - 1 =


(27 + 1 023)(10) =


1 050(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;
  • 1 050 ÷ 2 = 525 + 0;
  • 525 ÷ 2 = 262 + 1;
  • 262 ÷ 2 = 131 + 0;
  • 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) =


1050(10) =


100 0001 1010(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. 000 0111 0110 0111 1010 1011 0001 0 0000 0000 0000 0000 0000 0000 =


0000 1110 1100 1111 0101 0110 0010 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 1010


Mantissa (52 bits) =
0000 1110 1100 1111 0101 0110 0010 0000 0000 0000 0000 0000 0000


The base ten decimal number 141 982 385 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0001 1010 - 0000 1110 1100 1111 0101 0110 0010 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