55.111 111 111 108 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 108 7(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
55.111 111 111 108 7(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: 0.111 111 111 108 7.

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) 0.111 111 111 108 7 × 2 = 0 + 0.222 222 222 217 4;
  • 2) 0.222 222 222 217 4 × 2 = 0 + 0.444 444 444 434 8;
  • 3) 0.444 444 444 434 8 × 2 = 0 + 0.888 888 888 869 6;
  • 4) 0.888 888 888 869 6 × 2 = 1 + 0.777 777 777 739 2;
  • 5) 0.777 777 777 739 2 × 2 = 1 + 0.555 555 555 478 4;
  • 6) 0.555 555 555 478 4 × 2 = 1 + 0.111 111 110 956 8;
  • 7) 0.111 111 110 956 8 × 2 = 0 + 0.222 222 221 913 6;
  • 8) 0.222 222 221 913 6 × 2 = 0 + 0.444 444 443 827 2;
  • 9) 0.444 444 443 827 2 × 2 = 0 + 0.888 888 887 654 4;
  • 10) 0.888 888 887 654 4 × 2 = 1 + 0.777 777 775 308 8;
  • 11) 0.777 777 775 308 8 × 2 = 1 + 0.555 555 550 617 6;
  • 12) 0.555 555 550 617 6 × 2 = 1 + 0.111 111 101 235 2;
  • 13) 0.111 111 101 235 2 × 2 = 0 + 0.222 222 202 470 4;
  • 14) 0.222 222 202 470 4 × 2 = 0 + 0.444 444 404 940 8;
  • 15) 0.444 444 404 940 8 × 2 = 0 + 0.888 888 809 881 6;
  • 16) 0.888 888 809 881 6 × 2 = 1 + 0.777 777 619 763 2;
  • 17) 0.777 777 619 763 2 × 2 = 1 + 0.555 555 239 526 4;
  • 18) 0.555 555 239 526 4 × 2 = 1 + 0.111 110 479 052 8;
  • 19) 0.111 110 479 052 8 × 2 = 0 + 0.222 220 958 105 6;
  • 20) 0.222 220 958 105 6 × 2 = 0 + 0.444 441 916 211 2;
  • 21) 0.444 441 916 211 2 × 2 = 0 + 0.888 883 832 422 4;
  • 22) 0.888 883 832 422 4 × 2 = 1 + 0.777 767 664 844 8;
  • 23) 0.777 767 664 844 8 × 2 = 1 + 0.555 535 329 689 6;
  • 24) 0.555 535 329 689 6 × 2 = 1 + 0.111 070 659 379 2;
  • 25) 0.111 070 659 379 2 × 2 = 0 + 0.222 141 318 758 4;
  • 26) 0.222 141 318 758 4 × 2 = 0 + 0.444 282 637 516 8;
  • 27) 0.444 282 637 516 8 × 2 = 0 + 0.888 565 275 033 6;
  • 28) 0.888 565 275 033 6 × 2 = 1 + 0.777 130 550 067 2;
  • 29) 0.777 130 550 067 2 × 2 = 1 + 0.554 261 100 134 4;
  • 30) 0.554 261 100 134 4 × 2 = 1 + 0.108 522 200 268 8;
  • 31) 0.108 522 200 268 8 × 2 = 0 + 0.217 044 400 537 6;
  • 32) 0.217 044 400 537 6 × 2 = 0 + 0.434 088 801 075 2;
  • 33) 0.434 088 801 075 2 × 2 = 0 + 0.868 177 602 150 4;
  • 34) 0.868 177 602 150 4 × 2 = 1 + 0.736 355 204 300 8;
  • 35) 0.736 355 204 300 8 × 2 = 1 + 0.472 710 408 601 6;
  • 36) 0.472 710 408 601 6 × 2 = 0 + 0.945 420 817 203 2;
  • 37) 0.945 420 817 203 2 × 2 = 1 + 0.890 841 634 406 4;
  • 38) 0.890 841 634 406 4 × 2 = 1 + 0.781 683 268 812 8;
  • 39) 0.781 683 268 812 8 × 2 = 1 + 0.563 366 537 625 6;
  • 40) 0.563 366 537 625 6 × 2 = 1 + 0.126 733 075 251 2;
  • 41) 0.126 733 075 251 2 × 2 = 0 + 0.253 466 150 502 4;
  • 42) 0.253 466 150 502 4 × 2 = 0 + 0.506 932 301 004 8;
  • 43) 0.506 932 301 004 8 × 2 = 1 + 0.013 864 602 009 6;
  • 44) 0.013 864 602 009 6 × 2 = 0 + 0.027 729 204 019 2;
  • 45) 0.027 729 204 019 2 × 2 = 0 + 0.055 458 408 038 4;
  • 46) 0.055 458 408 038 4 × 2 = 0 + 0.110 916 816 076 8;
  • 47) 0.110 916 816 076 8 × 2 = 0 + 0.221 833 632 153 6;
  • 48) 0.221 833 632 153 6 × 2 = 0 + 0.443 667 264 307 2;
  • 49) 0.443 667 264 307 2 × 2 = 0 + 0.887 334 528 614 4;
  • 50) 0.887 334 528 614 4 × 2 = 1 + 0.774 669 057 228 8;
  • 51) 0.774 669 057 228 8 × 2 = 1 + 0.549 338 114 457 6;
  • 52) 0.549 338 114 457 6 × 2 = 1 + 0.098 676 228 915 2;
  • 53) 0.098 676 228 915 2 × 2 = 0 + 0.197 352 457 830 4;

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:


0.111 111 111 108 7(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0110 1111 0010 0000 0111 0(2)

5. Positive number before normalization:

55.111 111 111 108 7(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1111 0010 0000 0111 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:


55.111 111 111 108 7(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1111 0010 0000 0111 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 1111 0010 0000 0111 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0111 1001 0000 0011 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 0111 1001 0000 0011 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 0111 1001 0000 00 1110 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0111 1001 0000


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 1001 0000


Decimal number 55.111 111 111 108 7 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 1001 0000

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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