64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 17.020 706 1 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 17.020 706 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: 17.
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;
  • 17 ÷ 2 = 8 + 1;
  • 8 ÷ 2 = 4 + 0;
  • 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.


17(10) =


1 0001(2)


3. Convert to binary (base 2) the fractional part: 0.020 706 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) 0.020 706 1 × 2 = 0 + 0.041 412 2;
  • 2) 0.041 412 2 × 2 = 0 + 0.082 824 4;
  • 3) 0.082 824 4 × 2 = 0 + 0.165 648 8;
  • 4) 0.165 648 8 × 2 = 0 + 0.331 297 6;
  • 5) 0.331 297 6 × 2 = 0 + 0.662 595 2;
  • 6) 0.662 595 2 × 2 = 1 + 0.325 190 4;
  • 7) 0.325 190 4 × 2 = 0 + 0.650 380 8;
  • 8) 0.650 380 8 × 2 = 1 + 0.300 761 6;
  • 9) 0.300 761 6 × 2 = 0 + 0.601 523 2;
  • 10) 0.601 523 2 × 2 = 1 + 0.203 046 4;
  • 11) 0.203 046 4 × 2 = 0 + 0.406 092 8;
  • 12) 0.406 092 8 × 2 = 0 + 0.812 185 6;
  • 13) 0.812 185 6 × 2 = 1 + 0.624 371 2;
  • 14) 0.624 371 2 × 2 = 1 + 0.248 742 4;
  • 15) 0.248 742 4 × 2 = 0 + 0.497 484 8;
  • 16) 0.497 484 8 × 2 = 0 + 0.994 969 6;
  • 17) 0.994 969 6 × 2 = 1 + 0.989 939 2;
  • 18) 0.989 939 2 × 2 = 1 + 0.979 878 4;
  • 19) 0.979 878 4 × 2 = 1 + 0.959 756 8;
  • 20) 0.959 756 8 × 2 = 1 + 0.919 513 6;
  • 21) 0.919 513 6 × 2 = 1 + 0.839 027 2;
  • 22) 0.839 027 2 × 2 = 1 + 0.678 054 4;
  • 23) 0.678 054 4 × 2 = 1 + 0.356 108 8;
  • 24) 0.356 108 8 × 2 = 0 + 0.712 217 6;
  • 25) 0.712 217 6 × 2 = 1 + 0.424 435 2;
  • 26) 0.424 435 2 × 2 = 0 + 0.848 870 4;
  • 27) 0.848 870 4 × 2 = 1 + 0.697 740 8;
  • 28) 0.697 740 8 × 2 = 1 + 0.395 481 6;
  • 29) 0.395 481 6 × 2 = 0 + 0.790 963 2;
  • 30) 0.790 963 2 × 2 = 1 + 0.581 926 4;
  • 31) 0.581 926 4 × 2 = 1 + 0.163 852 8;
  • 32) 0.163 852 8 × 2 = 0 + 0.327 705 6;
  • 33) 0.327 705 6 × 2 = 0 + 0.655 411 2;
  • 34) 0.655 411 2 × 2 = 1 + 0.310 822 4;
  • 35) 0.310 822 4 × 2 = 0 + 0.621 644 8;
  • 36) 0.621 644 8 × 2 = 1 + 0.243 289 6;
  • 37) 0.243 289 6 × 2 = 0 + 0.486 579 2;
  • 38) 0.486 579 2 × 2 = 0 + 0.973 158 4;
  • 39) 0.973 158 4 × 2 = 1 + 0.946 316 8;
  • 40) 0.946 316 8 × 2 = 1 + 0.892 633 6;
  • 41) 0.892 633 6 × 2 = 1 + 0.785 267 2;
  • 42) 0.785 267 2 × 2 = 1 + 0.570 534 4;
  • 43) 0.570 534 4 × 2 = 1 + 0.141 068 8;
  • 44) 0.141 068 8 × 2 = 0 + 0.282 137 6;
  • 45) 0.282 137 6 × 2 = 0 + 0.564 275 2;
  • 46) 0.564 275 2 × 2 = 1 + 0.128 550 4;
  • 47) 0.128 550 4 × 2 = 0 + 0.257 100 8;
  • 48) 0.257 100 8 × 2 = 0 + 0.514 201 6;
  • 49) 0.514 201 6 × 2 = 1 + 0.028 403 2;
  • 50) 0.028 403 2 × 2 = 0 + 0.056 806 4;
  • 51) 0.056 806 4 × 2 = 0 + 0.113 612 8;
  • 52) 0.113 612 8 × 2 = 0 + 0.227 225 6;
  • 53) 0.227 225 6 × 2 = 0 + 0.454 451 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:


0.020 706 1(10) =


0.0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0(2)


5. Positive number before normalization:

17.020 706 1(10) =


1 0001.0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the left, so that only one non zero digit remains to the left of it:


17.020 706 1(10) =


1 0001.0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0(2) =


1 0001.0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0(2) × 20 =


1.0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0(2) × 24


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): 4


Mantissa (not normalized):
1.0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1000 0


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


4 + 2(11-1) - 1 =


(4 + 1 023)(10) =


1 027(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;
  • 1 027 ÷ 2 = 513 + 1;
  • 513 ÷ 2 = 256 + 1;
  • 256 ÷ 2 = 128 + 0;
  • 128 ÷ 2 = 64 + 0;
  • 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) =


1027(10) =


100 0000 0011(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. 0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100 1 0000 =


0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100


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 0011


Mantissa (52 bits) =
0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100


The base ten decimal number 17.020 706 1 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0000 0011 - 0001 0000 0101 0100 1100 1111 1110 1011 0110 0101 0011 1110 0100

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