64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -9.75 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -9.75(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:

|-9.75| = 9.75

2. First, convert to binary (in base 2) the integer part: 9.
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;
  • 9 ÷ 2 = 4 + 1;
  • 4 ÷ 2 = 2 + 0;
  • 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.


9(10) =


1001(2)


4. Convert to binary (base 2) the fractional part: 0.75.

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) 0.75 × 2 = 1 + 0.5;
  • 2) 0.5 × 2 = 1 + 0;

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:


0.75(10) =


0.11(2)


6. Positive number before normalization:

9.75(10) =


1001.11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the left, so that only one non zero digit remains to the left of it:


9.75(10) =


1001.11(2) =


1001.11(2) × 20 =


1.0011 1(2) × 23


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): 3


Mantissa (not normalized):
1.0011 1


9. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


3 + 2(11-1) - 1 =


(3 + 1 023)(10) =


1 026(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;
  • 1 026 ÷ 2 = 513 + 0;
  • 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;

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) =


1026(10) =


100 0000 0010(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 adding the necessary number of zeros to the right.


Mantissa (normalized) =


1. 0 0111 000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 =


0011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000


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 0010


Mantissa (52 bits) =
0011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000


The base ten decimal number -9.75 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0000 0010 - 0011 1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

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