64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 9 223 372 036 854 775 295 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 9 223 372 036 854 775 295(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;
  • 9 223 372 036 854 775 295 ÷ 2 = 4 611 686 018 427 387 647 + 1;
  • 4 611 686 018 427 387 647 ÷ 2 = 2 305 843 009 213 693 823 + 1;
  • 2 305 843 009 213 693 823 ÷ 2 = 1 152 921 504 606 846 911 + 1;
  • 1 152 921 504 606 846 911 ÷ 2 = 576 460 752 303 423 455 + 1;
  • 576 460 752 303 423 455 ÷ 2 = 288 230 376 151 711 727 + 1;
  • 288 230 376 151 711 727 ÷ 2 = 144 115 188 075 855 863 + 1;
  • 144 115 188 075 855 863 ÷ 2 = 72 057 594 037 927 931 + 1;
  • 72 057 594 037 927 931 ÷ 2 = 36 028 797 018 963 965 + 1;
  • 36 028 797 018 963 965 ÷ 2 = 18 014 398 509 481 982 + 1;
  • 18 014 398 509 481 982 ÷ 2 = 9 007 199 254 740 991 + 0;
  • 9 007 199 254 740 991 ÷ 2 = 4 503 599 627 370 495 + 1;
  • 4 503 599 627 370 495 ÷ 2 = 2 251 799 813 685 247 + 1;
  • 2 251 799 813 685 247 ÷ 2 = 1 125 899 906 842 623 + 1;
  • 1 125 899 906 842 623 ÷ 2 = 562 949 953 421 311 + 1;
  • 562 949 953 421 311 ÷ 2 = 281 474 976 710 655 + 1;
  • 281 474 976 710 655 ÷ 2 = 140 737 488 355 327 + 1;
  • 140 737 488 355 327 ÷ 2 = 70 368 744 177 663 + 1;
  • 70 368 744 177 663 ÷ 2 = 35 184 372 088 831 + 1;
  • 35 184 372 088 831 ÷ 2 = 17 592 186 044 415 + 1;
  • 17 592 186 044 415 ÷ 2 = 8 796 093 022 207 + 1;
  • 8 796 093 022 207 ÷ 2 = 4 398 046 511 103 + 1;
  • 4 398 046 511 103 ÷ 2 = 2 199 023 255 551 + 1;
  • 2 199 023 255 551 ÷ 2 = 1 099 511 627 775 + 1;
  • 1 099 511 627 775 ÷ 2 = 549 755 813 887 + 1;
  • 549 755 813 887 ÷ 2 = 274 877 906 943 + 1;
  • 274 877 906 943 ÷ 2 = 137 438 953 471 + 1;
  • 137 438 953 471 ÷ 2 = 68 719 476 735 + 1;
  • 68 719 476 735 ÷ 2 = 34 359 738 367 + 1;
  • 34 359 738 367 ÷ 2 = 17 179 869 183 + 1;
  • 17 179 869 183 ÷ 2 = 8 589 934 591 + 1;
  • 8 589 934 591 ÷ 2 = 4 294 967 295 + 1;
  • 4 294 967 295 ÷ 2 = 2 147 483 647 + 1;
  • 2 147 483 647 ÷ 2 = 1 073 741 823 + 1;
  • 1 073 741 823 ÷ 2 = 536 870 911 + 1;
  • 536 870 911 ÷ 2 = 268 435 455 + 1;
  • 268 435 455 ÷ 2 = 134 217 727 + 1;
  • 134 217 727 ÷ 2 = 67 108 863 + 1;
  • 67 108 863 ÷ 2 = 33 554 431 + 1;
  • 33 554 431 ÷ 2 = 16 777 215 + 1;
  • 16 777 215 ÷ 2 = 8 388 607 + 1;
  • 8 388 607 ÷ 2 = 4 194 303 + 1;
  • 4 194 303 ÷ 2 = 2 097 151 + 1;
  • 2 097 151 ÷ 2 = 1 048 575 + 1;
  • 1 048 575 ÷ 2 = 524 287 + 1;
  • 524 287 ÷ 2 = 262 143 + 1;
  • 262 143 ÷ 2 = 131 071 + 1;
  • 131 071 ÷ 2 = 65 535 + 1;
  • 65 535 ÷ 2 = 32 767 + 1;
  • 32 767 ÷ 2 = 16 383 + 1;
  • 16 383 ÷ 2 = 8 191 + 1;
  • 8 191 ÷ 2 = 4 095 + 1;
  • 4 095 ÷ 2 = 2 047 + 1;
  • 2 047 ÷ 2 = 1 023 + 1;
  • 1 023 ÷ 2 = 511 + 1;
  • 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;

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

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


9 223 372 036 854 775 295(10) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1111(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 62 positions to the left, so that only one non zero digit remains to the left of it:


9 223 372 036 854 775 295(10) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1111(2) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1111(2) × 20 =


1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0111 1111 11(2) × 262


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): 62


Mantissa (not normalized):
1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0111 1111 11


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =


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


62 + 2(11-1) - 1 =


(62 + 1 023)(10) =


1 085(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;
  • 1 085 ÷ 2 = 542 + 1;
  • 542 ÷ 2 = 271 + 0;
  • 271 ÷ 2 = 135 + 1;
  • 135 ÷ 2 = 67 + 1;
  • 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) =


1085(10) =


100 0011 1101(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 removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =


1. 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 01 1111 1111 =


1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


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 1101


Mantissa (52 bits) =
1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111


The base ten decimal number 9 223 372 036 854 775 295 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0011 1101 - 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111

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