64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: -195 577 743 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number -195 577 743(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:

|-195 577 743| = 195 577 743

2. 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;
  • 195 577 743 ÷ 2 = 97 788 871 + 1;
  • 97 788 871 ÷ 2 = 48 894 435 + 1;
  • 48 894 435 ÷ 2 = 24 447 217 + 1;
  • 24 447 217 ÷ 2 = 12 223 608 + 1;
  • 12 223 608 ÷ 2 = 6 111 804 + 0;
  • 6 111 804 ÷ 2 = 3 055 902 + 0;
  • 3 055 902 ÷ 2 = 1 527 951 + 0;
  • 1 527 951 ÷ 2 = 763 975 + 1;
  • 763 975 ÷ 2 = 381 987 + 1;
  • 381 987 ÷ 2 = 190 993 + 1;
  • 190 993 ÷ 2 = 95 496 + 1;
  • 95 496 ÷ 2 = 47 748 + 0;
  • 47 748 ÷ 2 = 23 874 + 0;
  • 23 874 ÷ 2 = 11 937 + 0;
  • 11 937 ÷ 2 = 5 968 + 1;
  • 5 968 ÷ 2 = 2 984 + 0;
  • 2 984 ÷ 2 = 1 492 + 0;
  • 1 492 ÷ 2 = 746 + 0;
  • 746 ÷ 2 = 373 + 0;
  • 373 ÷ 2 = 186 + 1;
  • 186 ÷ 2 = 93 + 0;
  • 93 ÷ 2 = 46 + 1;
  • 46 ÷ 2 = 23 + 0;
  • 23 ÷ 2 = 11 + 1;
  • 11 ÷ 2 = 5 + 1;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

3. Construct the base 2 representation of the positive number.

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


195 577 743(10) =


1011 1010 1000 0100 0111 1000 1111(2)


4. Normalize the binary representation of the number.

Shift the decimal mark 27 positions to the left, so that only one non zero digit remains to the left of it:


195 577 743(10) =


1011 1010 1000 0100 0111 1000 1111(2) =


1011 1010 1000 0100 0111 1000 1111(2) × 20 =


1.0111 0101 0000 1000 1111 0001 111(2) × 227


5. 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): 27


Mantissa (not normalized):
1.0111 0101 0000 1000 1111 0001 111


6. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


27 + 2(11-1) - 1 =


(27 + 1 023)(10) =


1 050(10)


7. 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;
  • 1 050 ÷ 2 = 525 + 0;
  • 525 ÷ 2 = 262 + 1;
  • 262 ÷ 2 = 131 + 0;
  • 131 ÷ 2 = 65 + 1;
  • 65 ÷ 2 = 32 + 1;
  • 32 ÷ 2 = 16 + 0;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1050(10) =


100 0001 1010(2)


9. 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. 011 1010 1000 0100 0111 1000 1111 0 0000 0000 0000 0000 0000 0000 =


0111 0101 0000 1000 1111 0001 1110 0000 0000 0000 0000 0000 0000


10. 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 0001 1010


Mantissa (52 bits) =
0111 0101 0000 1000 1111 0001 1110 0000 0000 0000 0000 0000 0000


The base ten decimal number -195 577 743 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
1 - 100 0001 1010 - 0111 0101 0000 1000 1111 0001 1110 0000 0000 0000 0000 0000 0000

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