Convert the Number -0.825 to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number. Detailed Explanations

Number -0.825(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)

The first steps we'll go through to make the conversion:

Convert to binary (to base 2) the integer part of the number.

Convert to binary the fractional part of the number.


1. Start with the positive version of the number:

|-0.825| = 0.825

2. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.


0(10) =


0(2)


4. Convert to binary (base 2) the fractional part: 0.825.

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) 0.825 × 2 = 1 + 0.65;
  • 2) 0.65 × 2 = 1 + 0.3;
  • 3) 0.3 × 2 = 0 + 0.6;
  • 4) 0.6 × 2 = 1 + 0.2;
  • 5) 0.2 × 2 = 0 + 0.4;
  • 6) 0.4 × 2 = 0 + 0.8;
  • 7) 0.8 × 2 = 1 + 0.6;
  • 8) 0.6 × 2 = 1 + 0.2;
  • 9) 0.2 × 2 = 0 + 0.4;
  • 10) 0.4 × 2 = 0 + 0.8;
  • 11) 0.8 × 2 = 1 + 0.6;
  • 12) 0.6 × 2 = 1 + 0.2;
  • 13) 0.2 × 2 = 0 + 0.4;
  • 14) 0.4 × 2 = 0 + 0.8;
  • 15) 0.8 × 2 = 1 + 0.6;
  • 16) 0.6 × 2 = 1 + 0.2;
  • 17) 0.2 × 2 = 0 + 0.4;
  • 18) 0.4 × 2 = 0 + 0.8;
  • 19) 0.8 × 2 = 1 + 0.6;
  • 20) 0.6 × 2 = 1 + 0.2;
  • 21) 0.2 × 2 = 0 + 0.4;
  • 22) 0.4 × 2 = 0 + 0.8;
  • 23) 0.8 × 2 = 1 + 0.6;
  • 24) 0.6 × 2 = 1 + 0.2;
  • 25) 0.2 × 2 = 0 + 0.4;
  • 26) 0.4 × 2 = 0 + 0.8;
  • 27) 0.8 × 2 = 1 + 0.6;
  • 28) 0.6 × 2 = 1 + 0.2;
  • 29) 0.2 × 2 = 0 + 0.4;
  • 30) 0.4 × 2 = 0 + 0.8;
  • 31) 0.8 × 2 = 1 + 0.6;
  • 32) 0.6 × 2 = 1 + 0.2;
  • 33) 0.2 × 2 = 0 + 0.4;
  • 34) 0.4 × 2 = 0 + 0.8;
  • 35) 0.8 × 2 = 1 + 0.6;
  • 36) 0.6 × 2 = 1 + 0.2;
  • 37) 0.2 × 2 = 0 + 0.4;
  • 38) 0.4 × 2 = 0 + 0.8;
  • 39) 0.8 × 2 = 1 + 0.6;
  • 40) 0.6 × 2 = 1 + 0.2;
  • 41) 0.2 × 2 = 0 + 0.4;
  • 42) 0.4 × 2 = 0 + 0.8;
  • 43) 0.8 × 2 = 1 + 0.6;
  • 44) 0.6 × 2 = 1 + 0.2;
  • 45) 0.2 × 2 = 0 + 0.4;
  • 46) 0.4 × 2 = 0 + 0.8;
  • 47) 0.8 × 2 = 1 + 0.6;
  • 48) 0.6 × 2 = 1 + 0.2;
  • 49) 0.2 × 2 = 0 + 0.4;
  • 50) 0.4 × 2 = 0 + 0.8;
  • 51) 0.8 × 2 = 1 + 0.6;
  • 52) 0.6 × 2 = 1 + 0.2;
  • 53) 0.2 × 2 = 0 + 0.4;

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:


0.825(10) =


0.1101 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0(2)


6. Positive number before normalization:

0.825(10) =


0.1101 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0(2)


The last steps we'll go through to make the conversion:

Normalize the binary representation of the number.

Adjust the exponent.

Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Normalize the mantissa.


7. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.825(10) =


0.1101 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0(2) =


0.1101 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0011 0(2) × 20 =


1.1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110(2) × 2-1


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): -1


Mantissa (not normalized):
1.1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


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


(-1 + 1 023)(10) =


1 022(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;
  • 1 022 ÷ 2 = 511 + 0;
  • 511 ÷ 2 = 255 + 1;
  • 255 ÷ 2 = 127 + 1;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =


1022(10) =


011 1111 1110(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, only if necessary (not the case here).


Mantissa (normalized) =


1. 1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 =


1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 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) =
011 1111 1110


Mantissa (52 bits) =
1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110


The base ten decimal number -0.825 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 011 1111 1110 - 1010 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110 0110

(64 bits IEEE 754)

Number -0.826 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Number -0.824 converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point representation = ?

Convert to 64 bit double precision IEEE 754 binary floating point representation standard

A number in 64 bit double precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (11 bits), mantissa (52 bits)

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Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

Available Base Conversions Between Decimal and Binary Systems

Conversions Between Decimal System Numbers (Written in Base Ten) and Binary System Numbers (Base Two and Computer Representation):


1. Integer -> Binary

2. Decimal -> Binary

3. Binary -> Integer

4. Binary -> Decimal