64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 39 405.1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 39 405.1(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. First, convert to binary (in base 2) the integer part: 39 405.
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;
  • 39 405 ÷ 2 = 19 702 + 1;
  • 19 702 ÷ 2 = 9 851 + 0;
  • 9 851 ÷ 2 = 4 925 + 1;
  • 4 925 ÷ 2 = 2 462 + 1;
  • 2 462 ÷ 2 = 1 231 + 0;
  • 1 231 ÷ 2 = 615 + 1;
  • 615 ÷ 2 = 307 + 1;
  • 307 ÷ 2 = 153 + 1;
  • 153 ÷ 2 = 76 + 1;
  • 76 ÷ 2 = 38 + 0;
  • 38 ÷ 2 = 19 + 0;
  • 19 ÷ 2 = 9 + 1;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

2. 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.


39 405(10) =


1001 1001 1110 1101(2)


3. Convert to binary (base 2) the fractional part: 0.1.

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) 0.1 × 2 = 0 + 0.2;
  • 2) 0.2 × 2 = 0 + 0.4;
  • 3) 0.4 × 2 = 0 + 0.8;
  • 4) 0.8 × 2 = 1 + 0.6;
  • 5) 0.6 × 2 = 1 + 0.2;
  • 6) 0.2 × 2 = 0 + 0.4;
  • 7) 0.4 × 2 = 0 + 0.8;
  • 8) 0.8 × 2 = 1 + 0.6;
  • 9) 0.6 × 2 = 1 + 0.2;
  • 10) 0.2 × 2 = 0 + 0.4;
  • 11) 0.4 × 2 = 0 + 0.8;
  • 12) 0.8 × 2 = 1 + 0.6;
  • 13) 0.6 × 2 = 1 + 0.2;
  • 14) 0.2 × 2 = 0 + 0.4;
  • 15) 0.4 × 2 = 0 + 0.8;
  • 16) 0.8 × 2 = 1 + 0.6;
  • 17) 0.6 × 2 = 1 + 0.2;
  • 18) 0.2 × 2 = 0 + 0.4;
  • 19) 0.4 × 2 = 0 + 0.8;
  • 20) 0.8 × 2 = 1 + 0.6;
  • 21) 0.6 × 2 = 1 + 0.2;
  • 22) 0.2 × 2 = 0 + 0.4;
  • 23) 0.4 × 2 = 0 + 0.8;
  • 24) 0.8 × 2 = 1 + 0.6;
  • 25) 0.6 × 2 = 1 + 0.2;
  • 26) 0.2 × 2 = 0 + 0.4;
  • 27) 0.4 × 2 = 0 + 0.8;
  • 28) 0.8 × 2 = 1 + 0.6;
  • 29) 0.6 × 2 = 1 + 0.2;
  • 30) 0.2 × 2 = 0 + 0.4;
  • 31) 0.4 × 2 = 0 + 0.8;
  • 32) 0.8 × 2 = 1 + 0.6;
  • 33) 0.6 × 2 = 1 + 0.2;
  • 34) 0.2 × 2 = 0 + 0.4;
  • 35) 0.4 × 2 = 0 + 0.8;
  • 36) 0.8 × 2 = 1 + 0.6;
  • 37) 0.6 × 2 = 1 + 0.2;
  • 38) 0.2 × 2 = 0 + 0.4;
  • 39) 0.4 × 2 = 0 + 0.8;
  • 40) 0.8 × 2 = 1 + 0.6;
  • 41) 0.6 × 2 = 1 + 0.2;
  • 42) 0.2 × 2 = 0 + 0.4;
  • 43) 0.4 × 2 = 0 + 0.8;
  • 44) 0.8 × 2 = 1 + 0.6;
  • 45) 0.6 × 2 = 1 + 0.2;
  • 46) 0.2 × 2 = 0 + 0.4;
  • 47) 0.4 × 2 = 0 + 0.8;
  • 48) 0.8 × 2 = 1 + 0.6;
  • 49) 0.6 × 2 = 1 + 0.2;
  • 50) 0.2 × 2 = 0 + 0.4;
  • 51) 0.4 × 2 = 0 + 0.8;
  • 52) 0.8 × 2 = 1 + 0.6;
  • 53) 0.6 × 2 = 1 + 0.2;

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...)


4. 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:


0.1(10) =


0.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1(2)


5. Positive number before normalization:

39 405.1(10) =


1001 1001 1110 1101.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 15 positions to the left, so that only one non zero digit remains to the left of it:


39 405.1(10) =


1001 1001 1110 1101.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1(2) =


1001 1001 1110 1101.0001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1001 1(2) × 20 =


1.0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011(2) × 215


7. 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): 15


Mantissa (not normalized):
1.0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


15 + 2(11-1) - 1 =


(15 + 1 023)(10) =


1 038(10)


9. 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;
  • 1 038 ÷ 2 = 519 + 0;
  • 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;

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

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


Exponent (adjusted) =


1038(10) =


100 0000 1110(2)


11. 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 removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 =


0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011


12. 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 1110


Mantissa (52 bits) =
0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011


The base ten decimal number 39 405.1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1110 - 0011 0011 1101 1010 0011 0011 0011 0011 0011 0011 0011 0011 0011

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