55.111 111 111 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 9(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 9(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 9.

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 9 × 2 = 0 + 0.222 222 223 8;
  • 2) 0.222 222 223 8 × 2 = 0 + 0.444 444 447 6;
  • 3) 0.444 444 447 6 × 2 = 0 + 0.888 888 895 2;
  • 4) 0.888 888 895 2 × 2 = 1 + 0.777 777 790 4;
  • 5) 0.777 777 790 4 × 2 = 1 + 0.555 555 580 8;
  • 6) 0.555 555 580 8 × 2 = 1 + 0.111 111 161 6;
  • 7) 0.111 111 161 6 × 2 = 0 + 0.222 222 323 2;
  • 8) 0.222 222 323 2 × 2 = 0 + 0.444 444 646 4;
  • 9) 0.444 444 646 4 × 2 = 0 + 0.888 889 292 8;
  • 10) 0.888 889 292 8 × 2 = 1 + 0.777 778 585 6;
  • 11) 0.777 778 585 6 × 2 = 1 + 0.555 557 171 2;
  • 12) 0.555 557 171 2 × 2 = 1 + 0.111 114 342 4;
  • 13) 0.111 114 342 4 × 2 = 0 + 0.222 228 684 8;
  • 14) 0.222 228 684 8 × 2 = 0 + 0.444 457 369 6;
  • 15) 0.444 457 369 6 × 2 = 0 + 0.888 914 739 2;
  • 16) 0.888 914 739 2 × 2 = 1 + 0.777 829 478 4;
  • 17) 0.777 829 478 4 × 2 = 1 + 0.555 658 956 8;
  • 18) 0.555 658 956 8 × 2 = 1 + 0.111 317 913 6;
  • 19) 0.111 317 913 6 × 2 = 0 + 0.222 635 827 2;
  • 20) 0.222 635 827 2 × 2 = 0 + 0.445 271 654 4;
  • 21) 0.445 271 654 4 × 2 = 0 + 0.890 543 308 8;
  • 22) 0.890 543 308 8 × 2 = 1 + 0.781 086 617 6;
  • 23) 0.781 086 617 6 × 2 = 1 + 0.562 173 235 2;
  • 24) 0.562 173 235 2 × 2 = 1 + 0.124 346 470 4;
  • 25) 0.124 346 470 4 × 2 = 0 + 0.248 692 940 8;
  • 26) 0.248 692 940 8 × 2 = 0 + 0.497 385 881 6;
  • 27) 0.497 385 881 6 × 2 = 0 + 0.994 771 763 2;
  • 28) 0.994 771 763 2 × 2 = 1 + 0.989 543 526 4;
  • 29) 0.989 543 526 4 × 2 = 1 + 0.979 087 052 8;
  • 30) 0.979 087 052 8 × 2 = 1 + 0.958 174 105 6;
  • 31) 0.958 174 105 6 × 2 = 1 + 0.916 348 211 2;
  • 32) 0.916 348 211 2 × 2 = 1 + 0.832 696 422 4;
  • 33) 0.832 696 422 4 × 2 = 1 + 0.665 392 844 8;
  • 34) 0.665 392 844 8 × 2 = 1 + 0.330 785 689 6;
  • 35) 0.330 785 689 6 × 2 = 0 + 0.661 571 379 2;
  • 36) 0.661 571 379 2 × 2 = 1 + 0.323 142 758 4;
  • 37) 0.323 142 758 4 × 2 = 0 + 0.646 285 516 8;
  • 38) 0.646 285 516 8 × 2 = 1 + 0.292 571 033 6;
  • 39) 0.292 571 033 6 × 2 = 0 + 0.585 142 067 2;
  • 40) 0.585 142 067 2 × 2 = 1 + 0.170 284 134 4;
  • 41) 0.170 284 134 4 × 2 = 0 + 0.340 568 268 8;
  • 42) 0.340 568 268 8 × 2 = 0 + 0.681 136 537 6;
  • 43) 0.681 136 537 6 × 2 = 1 + 0.362 273 075 2;
  • 44) 0.362 273 075 2 × 2 = 0 + 0.724 546 150 4;
  • 45) 0.724 546 150 4 × 2 = 1 + 0.449 092 300 8;
  • 46) 0.449 092 300 8 × 2 = 0 + 0.898 184 601 6;
  • 47) 0.898 184 601 6 × 2 = 1 + 0.796 369 203 2;
  • 48) 0.796 369 203 2 × 2 = 1 + 0.592 738 406 4;
  • 49) 0.592 738 406 4 × 2 = 1 + 0.185 476 812 8;
  • 50) 0.185 476 812 8 × 2 = 0 + 0.370 953 625 6;
  • 51) 0.370 953 625 6 × 2 = 0 + 0.741 907 251 2;
  • 52) 0.741 907 251 2 × 2 = 1 + 0.483 814 502 4;
  • 53) 0.483 814 502 4 × 2 = 0 + 0.967 629 004 8;

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 9(10) =

0.0001 1100 0111 0001 1100 0111 0001 1111 1101 0101 0010 1011 1001 0(2)

5. Positive number before normalization:

55.111 111 111 9(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1111 1101 0101 0010 1011 1001 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 9(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1111 1101 0101 0010 1011 1001 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1111 1101 0101 0010 1011 1001 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1111 1110 1010 1001 0101 1100 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 1111 1110 1010 1001 0101 1100 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 1111 1110 1010 1001 0101 11 0010 =

1011 1000 1110 0011 1000 1110 0011 1000 1111 1110 1010 1001 0101


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 1111 1110 1010 1001 0101


Decimal number 55.111 111 111 9 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 100 0000 0100 - 1011 1000 1110 0011 1000 1110 0011 1000 1111 1110 1010 1001 0101

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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