55.111 111 111 107 8 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 107 8(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
55.111 111 111 107 8(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: 0.111 111 111 107 8.

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) 0.111 111 111 107 8 × 2 = 0 + 0.222 222 222 215 6;
  • 2) 0.222 222 222 215 6 × 2 = 0 + 0.444 444 444 431 2;
  • 3) 0.444 444 444 431 2 × 2 = 0 + 0.888 888 888 862 4;
  • 4) 0.888 888 888 862 4 × 2 = 1 + 0.777 777 777 724 8;
  • 5) 0.777 777 777 724 8 × 2 = 1 + 0.555 555 555 449 6;
  • 6) 0.555 555 555 449 6 × 2 = 1 + 0.111 111 110 899 2;
  • 7) 0.111 111 110 899 2 × 2 = 0 + 0.222 222 221 798 4;
  • 8) 0.222 222 221 798 4 × 2 = 0 + 0.444 444 443 596 8;
  • 9) 0.444 444 443 596 8 × 2 = 0 + 0.888 888 887 193 6;
  • 10) 0.888 888 887 193 6 × 2 = 1 + 0.777 777 774 387 2;
  • 11) 0.777 777 774 387 2 × 2 = 1 + 0.555 555 548 774 4;
  • 12) 0.555 555 548 774 4 × 2 = 1 + 0.111 111 097 548 8;
  • 13) 0.111 111 097 548 8 × 2 = 0 + 0.222 222 195 097 6;
  • 14) 0.222 222 195 097 6 × 2 = 0 + 0.444 444 390 195 2;
  • 15) 0.444 444 390 195 2 × 2 = 0 + 0.888 888 780 390 4;
  • 16) 0.888 888 780 390 4 × 2 = 1 + 0.777 777 560 780 8;
  • 17) 0.777 777 560 780 8 × 2 = 1 + 0.555 555 121 561 6;
  • 18) 0.555 555 121 561 6 × 2 = 1 + 0.111 110 243 123 2;
  • 19) 0.111 110 243 123 2 × 2 = 0 + 0.222 220 486 246 4;
  • 20) 0.222 220 486 246 4 × 2 = 0 + 0.444 440 972 492 8;
  • 21) 0.444 440 972 492 8 × 2 = 0 + 0.888 881 944 985 6;
  • 22) 0.888 881 944 985 6 × 2 = 1 + 0.777 763 889 971 2;
  • 23) 0.777 763 889 971 2 × 2 = 1 + 0.555 527 779 942 4;
  • 24) 0.555 527 779 942 4 × 2 = 1 + 0.111 055 559 884 8;
  • 25) 0.111 055 559 884 8 × 2 = 0 + 0.222 111 119 769 6;
  • 26) 0.222 111 119 769 6 × 2 = 0 + 0.444 222 239 539 2;
  • 27) 0.444 222 239 539 2 × 2 = 0 + 0.888 444 479 078 4;
  • 28) 0.888 444 479 078 4 × 2 = 1 + 0.776 888 958 156 8;
  • 29) 0.776 888 958 156 8 × 2 = 1 + 0.553 777 916 313 6;
  • 30) 0.553 777 916 313 6 × 2 = 1 + 0.107 555 832 627 2;
  • 31) 0.107 555 832 627 2 × 2 = 0 + 0.215 111 665 254 4;
  • 32) 0.215 111 665 254 4 × 2 = 0 + 0.430 223 330 508 8;
  • 33) 0.430 223 330 508 8 × 2 = 0 + 0.860 446 661 017 6;
  • 34) 0.860 446 661 017 6 × 2 = 1 + 0.720 893 322 035 2;
  • 35) 0.720 893 322 035 2 × 2 = 1 + 0.441 786 644 070 4;
  • 36) 0.441 786 644 070 4 × 2 = 0 + 0.883 573 288 140 8;
  • 37) 0.883 573 288 140 8 × 2 = 1 + 0.767 146 576 281 6;
  • 38) 0.767 146 576 281 6 × 2 = 1 + 0.534 293 152 563 2;
  • 39) 0.534 293 152 563 2 × 2 = 1 + 0.068 586 305 126 4;
  • 40) 0.068 586 305 126 4 × 2 = 0 + 0.137 172 610 252 8;
  • 41) 0.137 172 610 252 8 × 2 = 0 + 0.274 345 220 505 6;
  • 42) 0.274 345 220 505 6 × 2 = 0 + 0.548 690 441 011 2;
  • 43) 0.548 690 441 011 2 × 2 = 1 + 0.097 380 882 022 4;
  • 44) 0.097 380 882 022 4 × 2 = 0 + 0.194 761 764 044 8;
  • 45) 0.194 761 764 044 8 × 2 = 0 + 0.389 523 528 089 6;
  • 46) 0.389 523 528 089 6 × 2 = 0 + 0.779 047 056 179 2;
  • 47) 0.779 047 056 179 2 × 2 = 1 + 0.558 094 112 358 4;
  • 48) 0.558 094 112 358 4 × 2 = 1 + 0.116 188 224 716 8;
  • 49) 0.116 188 224 716 8 × 2 = 0 + 0.232 376 449 433 6;
  • 50) 0.232 376 449 433 6 × 2 = 0 + 0.464 752 898 867 2;
  • 51) 0.464 752 898 867 2 × 2 = 0 + 0.929 505 797 734 4;
  • 52) 0.929 505 797 734 4 × 2 = 1 + 0.859 011 595 468 8;
  • 53) 0.859 011 595 468 8 × 2 = 1 + 0.718 023 190 937 6;

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:


0.111 111 111 107 8(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0110 1110 0010 0011 0001 1(2)

5. Positive number before normalization:

55.111 111 111 107 8(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1110 0010 0011 0001 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:


55.111 111 111 107 8(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1110 0010 0011 0001 1(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1110 0010 0011 0001 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0111 0001 0001 1000 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 0111 0001 0001 1000 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 0111 0001 0001 10 0011 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0111 0001 0001


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 0111 0001 0001


Decimal number 55.111 111 111 107 8 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 0111 0001 0001

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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