64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 2 356 568 909 876 540 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 2 356 568 909 876 540(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;
  • 2 356 568 909 876 540 ÷ 2 = 1 178 284 454 938 270 + 0;
  • 1 178 284 454 938 270 ÷ 2 = 589 142 227 469 135 + 0;
  • 589 142 227 469 135 ÷ 2 = 294 571 113 734 567 + 1;
  • 294 571 113 734 567 ÷ 2 = 147 285 556 867 283 + 1;
  • 147 285 556 867 283 ÷ 2 = 73 642 778 433 641 + 1;
  • 73 642 778 433 641 ÷ 2 = 36 821 389 216 820 + 1;
  • 36 821 389 216 820 ÷ 2 = 18 410 694 608 410 + 0;
  • 18 410 694 608 410 ÷ 2 = 9 205 347 304 205 + 0;
  • 9 205 347 304 205 ÷ 2 = 4 602 673 652 102 + 1;
  • 4 602 673 652 102 ÷ 2 = 2 301 336 826 051 + 0;
  • 2 301 336 826 051 ÷ 2 = 1 150 668 413 025 + 1;
  • 1 150 668 413 025 ÷ 2 = 575 334 206 512 + 1;
  • 575 334 206 512 ÷ 2 = 287 667 103 256 + 0;
  • 287 667 103 256 ÷ 2 = 143 833 551 628 + 0;
  • 143 833 551 628 ÷ 2 = 71 916 775 814 + 0;
  • 71 916 775 814 ÷ 2 = 35 958 387 907 + 0;
  • 35 958 387 907 ÷ 2 = 17 979 193 953 + 1;
  • 17 979 193 953 ÷ 2 = 8 989 596 976 + 1;
  • 8 989 596 976 ÷ 2 = 4 494 798 488 + 0;
  • 4 494 798 488 ÷ 2 = 2 247 399 244 + 0;
  • 2 247 399 244 ÷ 2 = 1 123 699 622 + 0;
  • 1 123 699 622 ÷ 2 = 561 849 811 + 0;
  • 561 849 811 ÷ 2 = 280 924 905 + 1;
  • 280 924 905 ÷ 2 = 140 462 452 + 1;
  • 140 462 452 ÷ 2 = 70 231 226 + 0;
  • 70 231 226 ÷ 2 = 35 115 613 + 0;
  • 35 115 613 ÷ 2 = 17 557 806 + 1;
  • 17 557 806 ÷ 2 = 8 778 903 + 0;
  • 8 778 903 ÷ 2 = 4 389 451 + 1;
  • 4 389 451 ÷ 2 = 2 194 725 + 1;
  • 2 194 725 ÷ 2 = 1 097 362 + 1;
  • 1 097 362 ÷ 2 = 548 681 + 0;
  • 548 681 ÷ 2 = 274 340 + 1;
  • 274 340 ÷ 2 = 137 170 + 0;
  • 137 170 ÷ 2 = 68 585 + 0;
  • 68 585 ÷ 2 = 34 292 + 1;
  • 34 292 ÷ 2 = 17 146 + 0;
  • 17 146 ÷ 2 = 8 573 + 0;
  • 8 573 ÷ 2 = 4 286 + 1;
  • 4 286 ÷ 2 = 2 143 + 0;
  • 2 143 ÷ 2 = 1 071 + 1;
  • 1 071 ÷ 2 = 535 + 1;
  • 535 ÷ 2 = 267 + 1;
  • 267 ÷ 2 = 133 + 1;
  • 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;

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

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


2 356 568 909 876 540(10) =


1000 0101 1111 0100 1001 0111 0100 1100 0011 0000 1101 0011 1100(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 51 positions to the left, so that only one non zero digit remains to the left of it:


2 356 568 909 876 540(10) =


1000 0101 1111 0100 1001 0111 0100 1100 0011 0000 1101 0011 1100(2) =


1000 0101 1111 0100 1001 0111 0100 1100 0011 0000 1101 0011 1100(2) × 20 =


1.0000 1011 1110 1001 0010 1110 1001 1000 0110 0001 1010 0111 100(2) × 251


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): 51


Mantissa (not normalized):
1.0000 1011 1110 1001 0010 1110 1001 1000 0110 0001 1010 0111 100


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


51 + 2(11-1) - 1 =


(51 + 1 023)(10) =


1 074(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;
  • 1 074 ÷ 2 = 537 + 0;
  • 537 ÷ 2 = 268 + 1;
  • 268 ÷ 2 = 134 + 0;
  • 134 ÷ 2 = 67 + 0;
  • 67 ÷ 2 = 33 + 1;
  • 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) =


1074(10) =


100 0011 0010(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. 000 0101 1111 0100 1001 0111 0100 1100 0011 0000 1101 0011 1100 0 =


0000 1011 1110 1001 0010 1110 1001 1000 0110 0001 1010 0111 1000


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 0011 0010


Mantissa (52 bits) =
0000 1011 1110 1001 0010 1110 1001 1000 0110 0001 1010 0111 1000


The base ten decimal number 2 356 568 909 876 540 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0011 0010 - 0000 1011 1110 1001 0010 1110 1001 1000 0110 0001 1010 0111 1000

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