0.555 555 555 555 33 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.555 555 555 555 33(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.555 555 555 555 33(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

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

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.555 555 555 555 33.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.555 555 555 555 33 × 2 = 1 + 0.111 111 111 110 66;
  • 2) 0.111 111 111 110 66 × 2 = 0 + 0.222 222 222 221 32;
  • 3) 0.222 222 222 221 32 × 2 = 0 + 0.444 444 444 442 64;
  • 4) 0.444 444 444 442 64 × 2 = 0 + 0.888 888 888 885 28;
  • 5) 0.888 888 888 885 28 × 2 = 1 + 0.777 777 777 770 56;
  • 6) 0.777 777 777 770 56 × 2 = 1 + 0.555 555 555 541 12;
  • 7) 0.555 555 555 541 12 × 2 = 1 + 0.111 111 111 082 24;
  • 8) 0.111 111 111 082 24 × 2 = 0 + 0.222 222 222 164 48;
  • 9) 0.222 222 222 164 48 × 2 = 0 + 0.444 444 444 328 96;
  • 10) 0.444 444 444 328 96 × 2 = 0 + 0.888 888 888 657 92;
  • 11) 0.888 888 888 657 92 × 2 = 1 + 0.777 777 777 315 84;
  • 12) 0.777 777 777 315 84 × 2 = 1 + 0.555 555 554 631 68;
  • 13) 0.555 555 554 631 68 × 2 = 1 + 0.111 111 109 263 36;
  • 14) 0.111 111 109 263 36 × 2 = 0 + 0.222 222 218 526 72;
  • 15) 0.222 222 218 526 72 × 2 = 0 + 0.444 444 437 053 44;
  • 16) 0.444 444 437 053 44 × 2 = 0 + 0.888 888 874 106 88;
  • 17) 0.888 888 874 106 88 × 2 = 1 + 0.777 777 748 213 76;
  • 18) 0.777 777 748 213 76 × 2 = 1 + 0.555 555 496 427 52;
  • 19) 0.555 555 496 427 52 × 2 = 1 + 0.111 110 992 855 04;
  • 20) 0.111 110 992 855 04 × 2 = 0 + 0.222 221 985 710 08;
  • 21) 0.222 221 985 710 08 × 2 = 0 + 0.444 443 971 420 16;
  • 22) 0.444 443 971 420 16 × 2 = 0 + 0.888 887 942 840 32;
  • 23) 0.888 887 942 840 32 × 2 = 1 + 0.777 775 885 680 64;
  • 24) 0.777 775 885 680 64 × 2 = 1 + 0.555 551 771 361 28;
  • 25) 0.555 551 771 361 28 × 2 = 1 + 0.111 103 542 722 56;
  • 26) 0.111 103 542 722 56 × 2 = 0 + 0.222 207 085 445 12;
  • 27) 0.222 207 085 445 12 × 2 = 0 + 0.444 414 170 890 24;
  • 28) 0.444 414 170 890 24 × 2 = 0 + 0.888 828 341 780 48;
  • 29) 0.888 828 341 780 48 × 2 = 1 + 0.777 656 683 560 96;
  • 30) 0.777 656 683 560 96 × 2 = 1 + 0.555 313 367 121 92;
  • 31) 0.555 313 367 121 92 × 2 = 1 + 0.110 626 734 243 84;
  • 32) 0.110 626 734 243 84 × 2 = 0 + 0.221 253 468 487 68;
  • 33) 0.221 253 468 487 68 × 2 = 0 + 0.442 506 936 975 36;
  • 34) 0.442 506 936 975 36 × 2 = 0 + 0.885 013 873 950 72;
  • 35) 0.885 013 873 950 72 × 2 = 1 + 0.770 027 747 901 44;
  • 36) 0.770 027 747 901 44 × 2 = 1 + 0.540 055 495 802 88;
  • 37) 0.540 055 495 802 88 × 2 = 1 + 0.080 110 991 605 76;
  • 38) 0.080 110 991 605 76 × 2 = 0 + 0.160 221 983 211 52;
  • 39) 0.160 221 983 211 52 × 2 = 0 + 0.320 443 966 423 04;
  • 40) 0.320 443 966 423 04 × 2 = 0 + 0.640 887 932 846 08;
  • 41) 0.640 887 932 846 08 × 2 = 1 + 0.281 775 865 692 16;
  • 42) 0.281 775 865 692 16 × 2 = 0 + 0.563 551 731 384 32;
  • 43) 0.563 551 731 384 32 × 2 = 1 + 0.127 103 462 768 64;
  • 44) 0.127 103 462 768 64 × 2 = 0 + 0.254 206 925 537 28;
  • 45) 0.254 206 925 537 28 × 2 = 0 + 0.508 413 851 074 56;
  • 46) 0.508 413 851 074 56 × 2 = 1 + 0.016 827 702 149 12;
  • 47) 0.016 827 702 149 12 × 2 = 0 + 0.033 655 404 298 24;
  • 48) 0.033 655 404 298 24 × 2 = 0 + 0.067 310 808 596 48;
  • 49) 0.067 310 808 596 48 × 2 = 0 + 0.134 621 617 192 96;
  • 50) 0.134 621 617 192 96 × 2 = 0 + 0.269 243 234 385 92;
  • 51) 0.269 243 234 385 92 × 2 = 0 + 0.538 486 468 771 84;
  • 52) 0.538 486 468 771 84 × 2 = 1 + 0.076 972 937 543 68;
  • 53) 0.076 972 937 543 68 × 2 = 0 + 0.153 945 875 087 36;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.555 555 555 555 33(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1010 0100 0001 0(2)

5. Positive number before normalization:

0.555 555 555 555 33(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1010 0100 0001 0(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 1 positions to the right, so that only one non zero digit remains to the left of it:


0.555 555 555 555 33(10) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1010 0100 0001 0(2) =

0.1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1010 0100 0001 0(2) × 20 =

1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010(2) × 2-1


7. 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): -1

Mantissa (not normalized):
1.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

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

(-1 + 1 023)(10) =

1 022(10)


9. 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;
  • 1 022 ÷ 2 = 511 + 0;
  • 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;

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

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


Exponent (adjusted) =

1022(10) =

011 1111 1110(2)


11. 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, only if necessary (not the case here).


Mantissa (normalized) =

1. 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010 =

0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010


12. 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) =
011 1111 1110

Mantissa (52 bits) =
0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010


Decimal number 0.555 555 555 555 33 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1111 1110 - 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 0100 1000 0010

» Convert decimal 0.555 555 555 555 39 to 64 bit double precision IEEE 754 binary floating point representation standard

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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