64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 2 415 452 124 156 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 2 415 452 124 156(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. 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;
  • 2 415 452 124 156 ÷ 2 = 1 207 726 062 078 + 0;
  • 1 207 726 062 078 ÷ 2 = 603 863 031 039 + 0;
  • 603 863 031 039 ÷ 2 = 301 931 515 519 + 1;
  • 301 931 515 519 ÷ 2 = 150 965 757 759 + 1;
  • 150 965 757 759 ÷ 2 = 75 482 878 879 + 1;
  • 75 482 878 879 ÷ 2 = 37 741 439 439 + 1;
  • 37 741 439 439 ÷ 2 = 18 870 719 719 + 1;
  • 18 870 719 719 ÷ 2 = 9 435 359 859 + 1;
  • 9 435 359 859 ÷ 2 = 4 717 679 929 + 1;
  • 4 717 679 929 ÷ 2 = 2 358 839 964 + 1;
  • 2 358 839 964 ÷ 2 = 1 179 419 982 + 0;
  • 1 179 419 982 ÷ 2 = 589 709 991 + 0;
  • 589 709 991 ÷ 2 = 294 854 995 + 1;
  • 294 854 995 ÷ 2 = 147 427 497 + 1;
  • 147 427 497 ÷ 2 = 73 713 748 + 1;
  • 73 713 748 ÷ 2 = 36 856 874 + 0;
  • 36 856 874 ÷ 2 = 18 428 437 + 0;
  • 18 428 437 ÷ 2 = 9 214 218 + 1;
  • 9 214 218 ÷ 2 = 4 607 109 + 0;
  • 4 607 109 ÷ 2 = 2 303 554 + 1;
  • 2 303 554 ÷ 2 = 1 151 777 + 0;
  • 1 151 777 ÷ 2 = 575 888 + 1;
  • 575 888 ÷ 2 = 287 944 + 0;
  • 287 944 ÷ 2 = 143 972 + 0;
  • 143 972 ÷ 2 = 71 986 + 0;
  • 71 986 ÷ 2 = 35 993 + 0;
  • 35 993 ÷ 2 = 17 996 + 1;
  • 17 996 ÷ 2 = 8 998 + 0;
  • 8 998 ÷ 2 = 4 499 + 0;
  • 4 499 ÷ 2 = 2 249 + 1;
  • 2 249 ÷ 2 = 1 124 + 1;
  • 1 124 ÷ 2 = 562 + 0;
  • 562 ÷ 2 = 281 + 0;
  • 281 ÷ 2 = 140 + 1;
  • 140 ÷ 2 = 70 + 0;
  • 70 ÷ 2 = 35 + 0;
  • 35 ÷ 2 = 17 + 1;
  • 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 positive number.

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


2 415 452 124 156(10) =


10 0011 0010 0110 0100 0010 1010 0111 0011 1111 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 41 positions to the left, so that only one non zero digit remains to the left of it:


2 415 452 124 156(10) =


10 0011 0010 0110 0100 0010 1010 0111 0011 1111 1100(2) =


10 0011 0010 0110 0100 0010 1010 0111 0011 1111 1100(2) × 20 =


1.0001 1001 0011 0010 0001 0101 0011 1001 1111 1110 0(2) × 241


4. 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): 41


Mantissa (not normalized):
1.0001 1001 0011 0010 0001 0101 0011 1001 1111 1110 0


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


41 + 2(11-1) - 1 =


(41 + 1 023)(10) =


1 064(10)


6. 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;
  • 1 064 ÷ 2 = 532 + 0;
  • 532 ÷ 2 = 266 + 0;
  • 266 ÷ 2 = 133 + 0;
  • 133 ÷ 2 = 66 + 1;
  • 66 ÷ 2 = 33 + 0;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =


1064(10) =


100 0010 1000(2)


8. 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 0011 0010 0110 0100 0010 1010 0111 0011 1111 1100 000 0000 0000 =


0001 1001 0011 0010 0001 0101 0011 1001 1111 1110 0000 0000 0000


9. 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 0010 1000


Mantissa (52 bits) =
0001 1001 0011 0010 0001 0101 0011 1001 1111 1110 0000 0000 0000


The base ten decimal number 2 415 452 124 156 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0010 1000 - 0001 1001 0011 0010 0001 0101 0011 1001 1111 1110 0000 0000 0000

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