55.111 111 111 114 38 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 114 38(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
55.111 111 111 114 38(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: 55.
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;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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.

55(10) =

11 0111(2)


3. Convert to binary (base 2) the fractional part: 0.111 111 111 114 38.

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) 0.111 111 111 114 38 × 2 = 0 + 0.222 222 222 228 76;
  • 2) 0.222 222 222 228 76 × 2 = 0 + 0.444 444 444 457 52;
  • 3) 0.444 444 444 457 52 × 2 = 0 + 0.888 888 888 915 04;
  • 4) 0.888 888 888 915 04 × 2 = 1 + 0.777 777 777 830 08;
  • 5) 0.777 777 777 830 08 × 2 = 1 + 0.555 555 555 660 16;
  • 6) 0.555 555 555 660 16 × 2 = 1 + 0.111 111 111 320 32;
  • 7) 0.111 111 111 320 32 × 2 = 0 + 0.222 222 222 640 64;
  • 8) 0.222 222 222 640 64 × 2 = 0 + 0.444 444 445 281 28;
  • 9) 0.444 444 445 281 28 × 2 = 0 + 0.888 888 890 562 56;
  • 10) 0.888 888 890 562 56 × 2 = 1 + 0.777 777 781 125 12;
  • 11) 0.777 777 781 125 12 × 2 = 1 + 0.555 555 562 250 24;
  • 12) 0.555 555 562 250 24 × 2 = 1 + 0.111 111 124 500 48;
  • 13) 0.111 111 124 500 48 × 2 = 0 + 0.222 222 249 000 96;
  • 14) 0.222 222 249 000 96 × 2 = 0 + 0.444 444 498 001 92;
  • 15) 0.444 444 498 001 92 × 2 = 0 + 0.888 888 996 003 84;
  • 16) 0.888 888 996 003 84 × 2 = 1 + 0.777 777 992 007 68;
  • 17) 0.777 777 992 007 68 × 2 = 1 + 0.555 555 984 015 36;
  • 18) 0.555 555 984 015 36 × 2 = 1 + 0.111 111 968 030 72;
  • 19) 0.111 111 968 030 72 × 2 = 0 + 0.222 223 936 061 44;
  • 20) 0.222 223 936 061 44 × 2 = 0 + 0.444 447 872 122 88;
  • 21) 0.444 447 872 122 88 × 2 = 0 + 0.888 895 744 245 76;
  • 22) 0.888 895 744 245 76 × 2 = 1 + 0.777 791 488 491 52;
  • 23) 0.777 791 488 491 52 × 2 = 1 + 0.555 582 976 983 04;
  • 24) 0.555 582 976 983 04 × 2 = 1 + 0.111 165 953 966 08;
  • 25) 0.111 165 953 966 08 × 2 = 0 + 0.222 331 907 932 16;
  • 26) 0.222 331 907 932 16 × 2 = 0 + 0.444 663 815 864 32;
  • 27) 0.444 663 815 864 32 × 2 = 0 + 0.889 327 631 728 64;
  • 28) 0.889 327 631 728 64 × 2 = 1 + 0.778 655 263 457 28;
  • 29) 0.778 655 263 457 28 × 2 = 1 + 0.557 310 526 914 56;
  • 30) 0.557 310 526 914 56 × 2 = 1 + 0.114 621 053 829 12;
  • 31) 0.114 621 053 829 12 × 2 = 0 + 0.229 242 107 658 24;
  • 32) 0.229 242 107 658 24 × 2 = 0 + 0.458 484 215 316 48;
  • 33) 0.458 484 215 316 48 × 2 = 0 + 0.916 968 430 632 96;
  • 34) 0.916 968 430 632 96 × 2 = 1 + 0.833 936 861 265 92;
  • 35) 0.833 936 861 265 92 × 2 = 1 + 0.667 873 722 531 84;
  • 36) 0.667 873 722 531 84 × 2 = 1 + 0.335 747 445 063 68;
  • 37) 0.335 747 445 063 68 × 2 = 0 + 0.671 494 890 127 36;
  • 38) 0.671 494 890 127 36 × 2 = 1 + 0.342 989 780 254 72;
  • 39) 0.342 989 780 254 72 × 2 = 0 + 0.685 979 560 509 44;
  • 40) 0.685 979 560 509 44 × 2 = 1 + 0.371 959 121 018 88;
  • 41) 0.371 959 121 018 88 × 2 = 0 + 0.743 918 242 037 76;
  • 42) 0.743 918 242 037 76 × 2 = 1 + 0.487 836 484 075 52;
  • 43) 0.487 836 484 075 52 × 2 = 0 + 0.975 672 968 151 04;
  • 44) 0.975 672 968 151 04 × 2 = 1 + 0.951 345 936 302 08;
  • 45) 0.951 345 936 302 08 × 2 = 1 + 0.902 691 872 604 16;
  • 46) 0.902 691 872 604 16 × 2 = 1 + 0.805 383 745 208 32;
  • 47) 0.805 383 745 208 32 × 2 = 1 + 0.610 767 490 416 64;
  • 48) 0.610 767 490 416 64 × 2 = 1 + 0.221 534 980 833 28;
  • 49) 0.221 534 980 833 28 × 2 = 0 + 0.443 069 961 666 56;
  • 50) 0.443 069 961 666 56 × 2 = 0 + 0.886 139 923 333 12;
  • 51) 0.886 139 923 333 12 × 2 = 1 + 0.772 279 846 666 24;
  • 52) 0.772 279 846 666 24 × 2 = 1 + 0.544 559 693 332 48;
  • 53) 0.544 559 693 332 48 × 2 = 1 + 0.089 119 386 664 96;

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:


0.111 111 111 114 38(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 0101 1111 0011 1(2)

5. Positive number before normalization:

55.111 111 111 114 38(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 0101 1111 0011 1(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 5 positions to the left, so that only one non zero digit remains to the left of it:


55.111 111 111 114 38(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 0101 1111 0011 1(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 0101 1111 0011 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 1111 1001 11(2) × 25


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): 5

Mantissa (not normalized):
1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 1111 1001 11


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

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

5 + 2(11-1) - 1 =

(5 + 1 023)(10) =

1 028(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;
  • 1 028 ÷ 2 = 514 + 0;
  • 514 ÷ 2 = 257 + 0;
  • 257 ÷ 2 = 128 + 1;
  • 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;

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) =

1028(10) =

100 0000 0100(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, 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. 1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 1111 10 0111 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 1111


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) =
100 0000 0100

Mantissa (52 bits) =
1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 1111


Decimal number 55.111 111 111 114 38 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1010 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