64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 1 602.080 35 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 1 602.080 35(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: 1 602.
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;
  • 1 602 ÷ 2 = 801 + 0;
  • 801 ÷ 2 = 400 + 1;
  • 400 ÷ 2 = 200 + 0;
  • 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 integer part of the number.

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


1 602(10) =


110 0100 0010(2)


3. Convert to binary (base 2) the fractional part: 0.080 35.

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) 0.080 35 × 2 = 0 + 0.160 7;
  • 2) 0.160 7 × 2 = 0 + 0.321 4;
  • 3) 0.321 4 × 2 = 0 + 0.642 8;
  • 4) 0.642 8 × 2 = 1 + 0.285 6;
  • 5) 0.285 6 × 2 = 0 + 0.571 2;
  • 6) 0.571 2 × 2 = 1 + 0.142 4;
  • 7) 0.142 4 × 2 = 0 + 0.284 8;
  • 8) 0.284 8 × 2 = 0 + 0.569 6;
  • 9) 0.569 6 × 2 = 1 + 0.139 2;
  • 10) 0.139 2 × 2 = 0 + 0.278 4;
  • 11) 0.278 4 × 2 = 0 + 0.556 8;
  • 12) 0.556 8 × 2 = 1 + 0.113 6;
  • 13) 0.113 6 × 2 = 0 + 0.227 2;
  • 14) 0.227 2 × 2 = 0 + 0.454 4;
  • 15) 0.454 4 × 2 = 0 + 0.908 8;
  • 16) 0.908 8 × 2 = 1 + 0.817 6;
  • 17) 0.817 6 × 2 = 1 + 0.635 2;
  • 18) 0.635 2 × 2 = 1 + 0.270 4;
  • 19) 0.270 4 × 2 = 0 + 0.540 8;
  • 20) 0.540 8 × 2 = 1 + 0.081 6;
  • 21) 0.081 6 × 2 = 0 + 0.163 2;
  • 22) 0.163 2 × 2 = 0 + 0.326 4;
  • 23) 0.326 4 × 2 = 0 + 0.652 8;
  • 24) 0.652 8 × 2 = 1 + 0.305 6;
  • 25) 0.305 6 × 2 = 0 + 0.611 2;
  • 26) 0.611 2 × 2 = 1 + 0.222 4;
  • 27) 0.222 4 × 2 = 0 + 0.444 8;
  • 28) 0.444 8 × 2 = 0 + 0.889 6;
  • 29) 0.889 6 × 2 = 1 + 0.779 2;
  • 30) 0.779 2 × 2 = 1 + 0.558 4;
  • 31) 0.558 4 × 2 = 1 + 0.116 8;
  • 32) 0.116 8 × 2 = 0 + 0.233 6;
  • 33) 0.233 6 × 2 = 0 + 0.467 2;
  • 34) 0.467 2 × 2 = 0 + 0.934 4;
  • 35) 0.934 4 × 2 = 1 + 0.868 8;
  • 36) 0.868 8 × 2 = 1 + 0.737 6;
  • 37) 0.737 6 × 2 = 1 + 0.475 2;
  • 38) 0.475 2 × 2 = 0 + 0.950 4;
  • 39) 0.950 4 × 2 = 1 + 0.900 8;
  • 40) 0.900 8 × 2 = 1 + 0.801 6;
  • 41) 0.801 6 × 2 = 1 + 0.603 2;
  • 42) 0.603 2 × 2 = 1 + 0.206 4;
  • 43) 0.206 4 × 2 = 0 + 0.412 8;
  • 44) 0.412 8 × 2 = 0 + 0.825 6;
  • 45) 0.825 6 × 2 = 1 + 0.651 2;
  • 46) 0.651 2 × 2 = 1 + 0.302 4;
  • 47) 0.302 4 × 2 = 0 + 0.604 8;
  • 48) 0.604 8 × 2 = 1 + 0.209 6;
  • 49) 0.209 6 × 2 = 0 + 0.419 2;
  • 50) 0.419 2 × 2 = 0 + 0.838 4;
  • 51) 0.838 4 × 2 = 1 + 0.676 8;
  • 52) 0.676 8 × 2 = 1 + 0.353 6;
  • 53) 0.353 6 × 2 = 0 + 0.707 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:


0.080 35(10) =


0.0001 0100 1001 0001 1101 0001 0100 1110 0011 1011 1100 1101 0011 0(2)


5. Positive number before normalization:

1 602.080 35(10) =


110 0100 0010.0001 0100 1001 0001 1101 0001 0100 1110 0011 1011 1100 1101 0011 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 10 positions to the left, so that only one non zero digit remains to the left of it:


1 602.080 35(10) =


110 0100 0010.0001 0100 1001 0001 1101 0001 0100 1110 0011 1011 1100 1101 0011 0(2) =


110 0100 0010.0001 0100 1001 0001 1101 0001 0100 1110 0011 1011 1100 1101 0011 0(2) × 20 =


1.1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111 0011 0100 110(2) × 210


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): 10


Mantissa (not normalized):
1.1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111 0011 0100 110


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


10 + 2(11-1) - 1 =


(10 + 1 023)(10) =


1 033(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;
  • 1 033 ÷ 2 = 516 + 1;
  • 516 ÷ 2 = 258 + 0;
  • 258 ÷ 2 = 129 + 0;
  • 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) =


1033(10) =


100 0000 1001(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. 1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111 001 1010 0110 =


1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111


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 1001


Mantissa (52 bits) =
1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111


The base ten decimal number 1 602.080 35 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 1001 - 1001 0000 1000 0101 0010 0100 0111 0100 0101 0011 1000 1110 1111

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