55.111 111 111 114 82 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 114 82(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 82(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 82.

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 82 × 2 = 0 + 0.222 222 222 229 64;
  • 2) 0.222 222 222 229 64 × 2 = 0 + 0.444 444 444 459 28;
  • 3) 0.444 444 444 459 28 × 2 = 0 + 0.888 888 888 918 56;
  • 4) 0.888 888 888 918 56 × 2 = 1 + 0.777 777 777 837 12;
  • 5) 0.777 777 777 837 12 × 2 = 1 + 0.555 555 555 674 24;
  • 6) 0.555 555 555 674 24 × 2 = 1 + 0.111 111 111 348 48;
  • 7) 0.111 111 111 348 48 × 2 = 0 + 0.222 222 222 696 96;
  • 8) 0.222 222 222 696 96 × 2 = 0 + 0.444 444 445 393 92;
  • 9) 0.444 444 445 393 92 × 2 = 0 + 0.888 888 890 787 84;
  • 10) 0.888 888 890 787 84 × 2 = 1 + 0.777 777 781 575 68;
  • 11) 0.777 777 781 575 68 × 2 = 1 + 0.555 555 563 151 36;
  • 12) 0.555 555 563 151 36 × 2 = 1 + 0.111 111 126 302 72;
  • 13) 0.111 111 126 302 72 × 2 = 0 + 0.222 222 252 605 44;
  • 14) 0.222 222 252 605 44 × 2 = 0 + 0.444 444 505 210 88;
  • 15) 0.444 444 505 210 88 × 2 = 0 + 0.888 889 010 421 76;
  • 16) 0.888 889 010 421 76 × 2 = 1 + 0.777 778 020 843 52;
  • 17) 0.777 778 020 843 52 × 2 = 1 + 0.555 556 041 687 04;
  • 18) 0.555 556 041 687 04 × 2 = 1 + 0.111 112 083 374 08;
  • 19) 0.111 112 083 374 08 × 2 = 0 + 0.222 224 166 748 16;
  • 20) 0.222 224 166 748 16 × 2 = 0 + 0.444 448 333 496 32;
  • 21) 0.444 448 333 496 32 × 2 = 0 + 0.888 896 666 992 64;
  • 22) 0.888 896 666 992 64 × 2 = 1 + 0.777 793 333 985 28;
  • 23) 0.777 793 333 985 28 × 2 = 1 + 0.555 586 667 970 56;
  • 24) 0.555 586 667 970 56 × 2 = 1 + 0.111 173 335 941 12;
  • 25) 0.111 173 335 941 12 × 2 = 0 + 0.222 346 671 882 24;
  • 26) 0.222 346 671 882 24 × 2 = 0 + 0.444 693 343 764 48;
  • 27) 0.444 693 343 764 48 × 2 = 0 + 0.889 386 687 528 96;
  • 28) 0.889 386 687 528 96 × 2 = 1 + 0.778 773 375 057 92;
  • 29) 0.778 773 375 057 92 × 2 = 1 + 0.557 546 750 115 84;
  • 30) 0.557 546 750 115 84 × 2 = 1 + 0.115 093 500 231 68;
  • 31) 0.115 093 500 231 68 × 2 = 0 + 0.230 187 000 463 36;
  • 32) 0.230 187 000 463 36 × 2 = 0 + 0.460 374 000 926 72;
  • 33) 0.460 374 000 926 72 × 2 = 0 + 0.920 748 001 853 44;
  • 34) 0.920 748 001 853 44 × 2 = 1 + 0.841 496 003 706 88;
  • 35) 0.841 496 003 706 88 × 2 = 1 + 0.682 992 007 413 76;
  • 36) 0.682 992 007 413 76 × 2 = 1 + 0.365 984 014 827 52;
  • 37) 0.365 984 014 827 52 × 2 = 0 + 0.731 968 029 655 04;
  • 38) 0.731 968 029 655 04 × 2 = 1 + 0.463 936 059 310 08;
  • 39) 0.463 936 059 310 08 × 2 = 0 + 0.927 872 118 620 16;
  • 40) 0.927 872 118 620 16 × 2 = 1 + 0.855 744 237 240 32;
  • 41) 0.855 744 237 240 32 × 2 = 1 + 0.711 488 474 480 64;
  • 42) 0.711 488 474 480 64 × 2 = 1 + 0.422 976 948 961 28;
  • 43) 0.422 976 948 961 28 × 2 = 0 + 0.845 953 897 922 56;
  • 44) 0.845 953 897 922 56 × 2 = 1 + 0.691 907 795 845 12;
  • 45) 0.691 907 795 845 12 × 2 = 1 + 0.383 815 591 690 24;
  • 46) 0.383 815 591 690 24 × 2 = 0 + 0.767 631 183 380 48;
  • 47) 0.767 631 183 380 48 × 2 = 1 + 0.535 262 366 760 96;
  • 48) 0.535 262 366 760 96 × 2 = 1 + 0.070 524 733 521 92;
  • 49) 0.070 524 733 521 92 × 2 = 0 + 0.141 049 467 043 84;
  • 50) 0.141 049 467 043 84 × 2 = 0 + 0.282 098 934 087 68;
  • 51) 0.282 098 934 087 68 × 2 = 0 + 0.564 197 868 175 36;
  • 52) 0.564 197 868 175 36 × 2 = 1 + 0.128 395 736 350 72;
  • 53) 0.128 395 736 350 72 × 2 = 0 + 0.256 791 472 701 44;

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 82(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 1101 1011 0001 0(2)

5. Positive number before normalization:

55.111 111 111 114 82(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 1101 1011 0001 0(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 82(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 1101 1011 0001 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0101 1101 1011 0001 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1110 1101 1000 10(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 1110 1101 1000 10


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 1110 1101 10 0010 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1010 1110 1101


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 1110 1101


Decimal number 55.111 111 111 114 82 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 1110 1101

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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