55.111 111 111 090 9 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 090 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
55.111 111 111 090 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: 0.111 111 111 090 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) 0.111 111 111 090 9 × 2 = 0 + 0.222 222 222 181 8;
  • 2) 0.222 222 222 181 8 × 2 = 0 + 0.444 444 444 363 6;
  • 3) 0.444 444 444 363 6 × 2 = 0 + 0.888 888 888 727 2;
  • 4) 0.888 888 888 727 2 × 2 = 1 + 0.777 777 777 454 4;
  • 5) 0.777 777 777 454 4 × 2 = 1 + 0.555 555 554 908 8;
  • 6) 0.555 555 554 908 8 × 2 = 1 + 0.111 111 109 817 6;
  • 7) 0.111 111 109 817 6 × 2 = 0 + 0.222 222 219 635 2;
  • 8) 0.222 222 219 635 2 × 2 = 0 + 0.444 444 439 270 4;
  • 9) 0.444 444 439 270 4 × 2 = 0 + 0.888 888 878 540 8;
  • 10) 0.888 888 878 540 8 × 2 = 1 + 0.777 777 757 081 6;
  • 11) 0.777 777 757 081 6 × 2 = 1 + 0.555 555 514 163 2;
  • 12) 0.555 555 514 163 2 × 2 = 1 + 0.111 111 028 326 4;
  • 13) 0.111 111 028 326 4 × 2 = 0 + 0.222 222 056 652 8;
  • 14) 0.222 222 056 652 8 × 2 = 0 + 0.444 444 113 305 6;
  • 15) 0.444 444 113 305 6 × 2 = 0 + 0.888 888 226 611 2;
  • 16) 0.888 888 226 611 2 × 2 = 1 + 0.777 776 453 222 4;
  • 17) 0.777 776 453 222 4 × 2 = 1 + 0.555 552 906 444 8;
  • 18) 0.555 552 906 444 8 × 2 = 1 + 0.111 105 812 889 6;
  • 19) 0.111 105 812 889 6 × 2 = 0 + 0.222 211 625 779 2;
  • 20) 0.222 211 625 779 2 × 2 = 0 + 0.444 423 251 558 4;
  • 21) 0.444 423 251 558 4 × 2 = 0 + 0.888 846 503 116 8;
  • 22) 0.888 846 503 116 8 × 2 = 1 + 0.777 693 006 233 6;
  • 23) 0.777 693 006 233 6 × 2 = 1 + 0.555 386 012 467 2;
  • 24) 0.555 386 012 467 2 × 2 = 1 + 0.110 772 024 934 4;
  • 25) 0.110 772 024 934 4 × 2 = 0 + 0.221 544 049 868 8;
  • 26) 0.221 544 049 868 8 × 2 = 0 + 0.443 088 099 737 6;
  • 27) 0.443 088 099 737 6 × 2 = 0 + 0.886 176 199 475 2;
  • 28) 0.886 176 199 475 2 × 2 = 1 + 0.772 352 398 950 4;
  • 29) 0.772 352 398 950 4 × 2 = 1 + 0.544 704 797 900 8;
  • 30) 0.544 704 797 900 8 × 2 = 1 + 0.089 409 595 801 6;
  • 31) 0.089 409 595 801 6 × 2 = 0 + 0.178 819 191 603 2;
  • 32) 0.178 819 191 603 2 × 2 = 0 + 0.357 638 383 206 4;
  • 33) 0.357 638 383 206 4 × 2 = 0 + 0.715 276 766 412 8;
  • 34) 0.715 276 766 412 8 × 2 = 1 + 0.430 553 532 825 6;
  • 35) 0.430 553 532 825 6 × 2 = 0 + 0.861 107 065 651 2;
  • 36) 0.861 107 065 651 2 × 2 = 1 + 0.722 214 131 302 4;
  • 37) 0.722 214 131 302 4 × 2 = 1 + 0.444 428 262 604 8;
  • 38) 0.444 428 262 604 8 × 2 = 0 + 0.888 856 525 209 6;
  • 39) 0.888 856 525 209 6 × 2 = 1 + 0.777 713 050 419 2;
  • 40) 0.777 713 050 419 2 × 2 = 1 + 0.555 426 100 838 4;
  • 41) 0.555 426 100 838 4 × 2 = 1 + 0.110 852 201 676 8;
  • 42) 0.110 852 201 676 8 × 2 = 0 + 0.221 704 403 353 6;
  • 43) 0.221 704 403 353 6 × 2 = 0 + 0.443 408 806 707 2;
  • 44) 0.443 408 806 707 2 × 2 = 0 + 0.886 817 613 414 4;
  • 45) 0.886 817 613 414 4 × 2 = 1 + 0.773 635 226 828 8;
  • 46) 0.773 635 226 828 8 × 2 = 1 + 0.547 270 453 657 6;
  • 47) 0.547 270 453 657 6 × 2 = 1 + 0.094 540 907 315 2;
  • 48) 0.094 540 907 315 2 × 2 = 0 + 0.189 081 814 630 4;
  • 49) 0.189 081 814 630 4 × 2 = 0 + 0.378 163 629 260 8;
  • 50) 0.378 163 629 260 8 × 2 = 0 + 0.756 327 258 521 6;
  • 51) 0.756 327 258 521 6 × 2 = 1 + 0.512 654 517 043 2;
  • 52) 0.512 654 517 043 2 × 2 = 1 + 0.025 309 034 086 4;
  • 53) 0.025 309 034 086 4 × 2 = 0 + 0.050 618 068 172 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:


0.111 111 111 090 9(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0101 1011 1000 1110 0011 0(2)

5. Positive number before normalization:

55.111 111 111 090 9(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1011 1000 1110 0011 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:


55.111 111 111 090 9(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1011 1000 1110 0011 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1011 1000 1110 0011 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1101 1100 0111 0001 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 0010 1101 1100 0111 0001 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 0010 1101 1100 0111 00 0110 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1101 1100 0111


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 0010 1101 1100 0111


Decimal number 55.111 111 111 090 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 1110 0010 1101 1100 0111

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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