64bit IEEE 754: Decimal ↗ Double Precision Floating Point Binary: 9 223 372 036 854 775 290 Convert the Number to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard, From a Base Ten Decimal System Number

Number 9 223 372 036 854 775 290(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;
  • 9 223 372 036 854 775 290 ÷ 2 = 4 611 686 018 427 387 645 + 0;
  • 4 611 686 018 427 387 645 ÷ 2 = 2 305 843 009 213 693 822 + 1;
  • 2 305 843 009 213 693 822 ÷ 2 = 1 152 921 504 606 846 911 + 0;
  • 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.


9 223 372 036 854 775 290(10) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1010(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:


9 223 372 036 854 775 290(10) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1010(2) =


111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1101 1111 1010(2) × 20 =


1.1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 1111 0111 1110 10(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 1110 10


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 1010 =


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 9 223 372 036 854 775 290 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

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. 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): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100