64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 9 223 372 036 854 775 840 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 9 223 372 036 854 775 840(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;
  • 9 223 372 036 854 775 840 ÷ 2 = 4 611 686 018 427 387 920 + 0;
  • 4 611 686 018 427 387 920 ÷ 2 = 2 305 843 009 213 693 960 + 0;
  • 2 305 843 009 213 693 960 ÷ 2 = 1 152 921 504 606 846 980 + 0;
  • 1 152 921 504 606 846 980 ÷ 2 = 576 460 752 303 423 490 + 0;
  • 576 460 752 303 423 490 ÷ 2 = 288 230 376 151 711 745 + 0;
  • 288 230 376 151 711 745 ÷ 2 = 144 115 188 075 855 872 + 1;
  • 144 115 188 075 855 872 ÷ 2 = 72 057 594 037 927 936 + 0;
  • 72 057 594 037 927 936 ÷ 2 = 36 028 797 018 963 968 + 0;
  • 36 028 797 018 963 968 ÷ 2 = 18 014 398 509 481 984 + 0;
  • 18 014 398 509 481 984 ÷ 2 = 9 007 199 254 740 992 + 0;
  • 9 007 199 254 740 992 ÷ 2 = 4 503 599 627 370 496 + 0;
  • 4 503 599 627 370 496 ÷ 2 = 2 251 799 813 685 248 + 0;
  • 2 251 799 813 685 248 ÷ 2 = 1 125 899 906 842 624 + 0;
  • 1 125 899 906 842 624 ÷ 2 = 562 949 953 421 312 + 0;
  • 562 949 953 421 312 ÷ 2 = 281 474 976 710 656 + 0;
  • 281 474 976 710 656 ÷ 2 = 140 737 488 355 328 + 0;
  • 140 737 488 355 328 ÷ 2 = 70 368 744 177 664 + 0;
  • 70 368 744 177 664 ÷ 2 = 35 184 372 088 832 + 0;
  • 35 184 372 088 832 ÷ 2 = 17 592 186 044 416 + 0;
  • 17 592 186 044 416 ÷ 2 = 8 796 093 022 208 + 0;
  • 8 796 093 022 208 ÷ 2 = 4 398 046 511 104 + 0;
  • 4 398 046 511 104 ÷ 2 = 2 199 023 255 552 + 0;
  • 2 199 023 255 552 ÷ 2 = 1 099 511 627 776 + 0;
  • 1 099 511 627 776 ÷ 2 = 549 755 813 888 + 0;
  • 549 755 813 888 ÷ 2 = 274 877 906 944 + 0;
  • 274 877 906 944 ÷ 2 = 137 438 953 472 + 0;
  • 137 438 953 472 ÷ 2 = 68 719 476 736 + 0;
  • 68 719 476 736 ÷ 2 = 34 359 738 368 + 0;
  • 34 359 738 368 ÷ 2 = 17 179 869 184 + 0;
  • 17 179 869 184 ÷ 2 = 8 589 934 592 + 0;
  • 8 589 934 592 ÷ 2 = 4 294 967 296 + 0;
  • 4 294 967 296 ÷ 2 = 2 147 483 648 + 0;
  • 2 147 483 648 ÷ 2 = 1 073 741 824 + 0;
  • 1 073 741 824 ÷ 2 = 536 870 912 + 0;
  • 536 870 912 ÷ 2 = 268 435 456 + 0;
  • 268 435 456 ÷ 2 = 134 217 728 + 0;
  • 134 217 728 ÷ 2 = 67 108 864 + 0;
  • 67 108 864 ÷ 2 = 33 554 432 + 0;
  • 33 554 432 ÷ 2 = 16 777 216 + 0;
  • 16 777 216 ÷ 2 = 8 388 608 + 0;
  • 8 388 608 ÷ 2 = 4 194 304 + 0;
  • 4 194 304 ÷ 2 = 2 097 152 + 0;
  • 2 097 152 ÷ 2 = 1 048 576 + 0;
  • 1 048 576 ÷ 2 = 524 288 + 0;
  • 524 288 ÷ 2 = 262 144 + 0;
  • 262 144 ÷ 2 = 131 072 + 0;
  • 131 072 ÷ 2 = 65 536 + 0;
  • 65 536 ÷ 2 = 32 768 + 0;
  • 32 768 ÷ 2 = 16 384 + 0;
  • 16 384 ÷ 2 = 8 192 + 0;
  • 8 192 ÷ 2 = 4 096 + 0;
  • 4 096 ÷ 2 = 2 048 + 0;
  • 2 048 ÷ 2 = 1 024 + 0;
  • 1 024 ÷ 2 = 512 + 0;
  • 512 ÷ 2 = 256 + 0;
  • 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;

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

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


9 223 372 036 854 775 840(10) =

1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000(2)


3. Normalize the binary representation of the number.

Shift the decimal mark 63 positions to the left, so that only one non zero digit remains to the left of it:


9 223 372 036 854 775 840(10) =

1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000(2) =

1000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0010 0000(2) × 20 =

1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 000(2) × 263


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): 63

Mantissa (not normalized):
1.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0100 000


5. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

63 + 2(11-1) - 1 =

(63 + 1 023)(10) =

1 086(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;
  • 1 086 ÷ 2 = 543 + 0;
  • 543 ÷ 2 = 271 + 1;
  • 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) =

1086(10) =

100 0011 1110(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. 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 000 0010 0000 =

0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 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 0011 1110

Mantissa (52 bits) =
0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000


The base ten decimal number 9 223 372 036 854 775 840 converted and written in 64 bit double precision IEEE 754 binary floating point representation:
0 - 100 0011 1110 - 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000

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