64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -160.207 02 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -160.207 02(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. Start with the positive version of the number:

|-160.207 02| = 160.207 02

2. First, convert to binary (in base 2) the integer part: 160.
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;
  • 160 ÷ 2 = 80 + 0;
  • 80 ÷ 2 = 40 + 0;
  • 40 ÷ 2 = 20 + 0;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. 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.


160(10) =


1010 0000(2)


4. Convert to binary (base 2) the fractional part: 0.207 02.

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) 0.207 02 × 2 = 0 + 0.414 04;
  • 2) 0.414 04 × 2 = 0 + 0.828 08;
  • 3) 0.828 08 × 2 = 1 + 0.656 16;
  • 4) 0.656 16 × 2 = 1 + 0.312 32;
  • 5) 0.312 32 × 2 = 0 + 0.624 64;
  • 6) 0.624 64 × 2 = 1 + 0.249 28;
  • 7) 0.249 28 × 2 = 0 + 0.498 56;
  • 8) 0.498 56 × 2 = 0 + 0.997 12;
  • 9) 0.997 12 × 2 = 1 + 0.994 24;
  • 10) 0.994 24 × 2 = 1 + 0.988 48;
  • 11) 0.988 48 × 2 = 1 + 0.976 96;
  • 12) 0.976 96 × 2 = 1 + 0.953 92;
  • 13) 0.953 92 × 2 = 1 + 0.907 84;
  • 14) 0.907 84 × 2 = 1 + 0.815 68;
  • 15) 0.815 68 × 2 = 1 + 0.631 36;
  • 16) 0.631 36 × 2 = 1 + 0.262 72;
  • 17) 0.262 72 × 2 = 0 + 0.525 44;
  • 18) 0.525 44 × 2 = 1 + 0.050 88;
  • 19) 0.050 88 × 2 = 0 + 0.101 76;
  • 20) 0.101 76 × 2 = 0 + 0.203 52;
  • 21) 0.203 52 × 2 = 0 + 0.407 04;
  • 22) 0.407 04 × 2 = 0 + 0.814 08;
  • 23) 0.814 08 × 2 = 1 + 0.628 16;
  • 24) 0.628 16 × 2 = 1 + 0.256 32;
  • 25) 0.256 32 × 2 = 0 + 0.512 64;
  • 26) 0.512 64 × 2 = 1 + 0.025 28;
  • 27) 0.025 28 × 2 = 0 + 0.050 56;
  • 28) 0.050 56 × 2 = 0 + 0.101 12;
  • 29) 0.101 12 × 2 = 0 + 0.202 24;
  • 30) 0.202 24 × 2 = 0 + 0.404 48;
  • 31) 0.404 48 × 2 = 0 + 0.808 96;
  • 32) 0.808 96 × 2 = 1 + 0.617 92;
  • 33) 0.617 92 × 2 = 1 + 0.235 84;
  • 34) 0.235 84 × 2 = 0 + 0.471 68;
  • 35) 0.471 68 × 2 = 0 + 0.943 36;
  • 36) 0.943 36 × 2 = 1 + 0.886 72;
  • 37) 0.886 72 × 2 = 1 + 0.773 44;
  • 38) 0.773 44 × 2 = 1 + 0.546 88;
  • 39) 0.546 88 × 2 = 1 + 0.093 76;
  • 40) 0.093 76 × 2 = 0 + 0.187 52;
  • 41) 0.187 52 × 2 = 0 + 0.375 04;
  • 42) 0.375 04 × 2 = 0 + 0.750 08;
  • 43) 0.750 08 × 2 = 1 + 0.500 16;
  • 44) 0.500 16 × 2 = 1 + 0.000 32;
  • 45) 0.000 32 × 2 = 0 + 0.000 64;
  • 46) 0.000 64 × 2 = 0 + 0.001 28;
  • 47) 0.001 28 × 2 = 0 + 0.002 56;
  • 48) 0.002 56 × 2 = 0 + 0.005 12;
  • 49) 0.005 12 × 2 = 0 + 0.010 24;
  • 50) 0.010 24 × 2 = 0 + 0.020 48;
  • 51) 0.020 48 × 2 = 0 + 0.040 96;
  • 52) 0.040 96 × 2 = 0 + 0.081 92;
  • 53) 0.081 92 × 2 = 0 + 0.163 84;

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...)


5. 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:


0.207 02(10) =


0.0011 0100 1111 1111 0100 0011 0100 0001 1001 1110 0011 0000 0000 0(2)


6. Positive number before normalization:

160.207 02(10) =


1010 0000.0011 0100 1111 1111 0100 0011 0100 0001 1001 1110 0011 0000 0000 0(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 7 positions to the left, so that only one non zero digit remains to the left of it:


160.207 02(10) =


1010 0000.0011 0100 1111 1111 0100 0011 0100 0001 1001 1110 0011 0000 0000 0(2) =


1010 0000.0011 0100 1111 1111 0100 0011 0100 0001 1001 1110 0011 0000 0000 0(2) × 20 =


1.0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110 0000 0000(2) × 27


8. 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 1 (a negative number)


Exponent (unadjusted): 7


Mantissa (not normalized):
1.0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110 0000 0000


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


7 + 2(11-1) - 1 =


(7 + 1 023)(10) =


1 030(10)


10. 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;
  • 1 030 ÷ 2 = 515 + 0;
  • 515 ÷ 2 = 257 + 1;
  • 257 ÷ 2 = 128 + 1;
  • 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;

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

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


Exponent (adjusted) =


1030(10) =


100 0000 0110(2)


12. 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. 0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110 0000 0000 =


0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110


13. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
1 (a negative number)


Exponent (11 bits) =
100 0000 0110


Mantissa (52 bits) =
0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110


The base ten decimal number -160.207 02 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0110 - 0100 0000 0110 1001 1111 1110 1000 0110 1000 0011 0011 1100 0110

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