55.111 111 111 092 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 092 3(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
55.111 111 111 092 3(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: 0.111 111 111 092 3.

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) 0.111 111 111 092 3 × 2 = 0 + 0.222 222 222 184 6;
  • 2) 0.222 222 222 184 6 × 2 = 0 + 0.444 444 444 369 2;
  • 3) 0.444 444 444 369 2 × 2 = 0 + 0.888 888 888 738 4;
  • 4) 0.888 888 888 738 4 × 2 = 1 + 0.777 777 777 476 8;
  • 5) 0.777 777 777 476 8 × 2 = 1 + 0.555 555 554 953 6;
  • 6) 0.555 555 554 953 6 × 2 = 1 + 0.111 111 109 907 2;
  • 7) 0.111 111 109 907 2 × 2 = 0 + 0.222 222 219 814 4;
  • 8) 0.222 222 219 814 4 × 2 = 0 + 0.444 444 439 628 8;
  • 9) 0.444 444 439 628 8 × 2 = 0 + 0.888 888 879 257 6;
  • 10) 0.888 888 879 257 6 × 2 = 1 + 0.777 777 758 515 2;
  • 11) 0.777 777 758 515 2 × 2 = 1 + 0.555 555 517 030 4;
  • 12) 0.555 555 517 030 4 × 2 = 1 + 0.111 111 034 060 8;
  • 13) 0.111 111 034 060 8 × 2 = 0 + 0.222 222 068 121 6;
  • 14) 0.222 222 068 121 6 × 2 = 0 + 0.444 444 136 243 2;
  • 15) 0.444 444 136 243 2 × 2 = 0 + 0.888 888 272 486 4;
  • 16) 0.888 888 272 486 4 × 2 = 1 + 0.777 776 544 972 8;
  • 17) 0.777 776 544 972 8 × 2 = 1 + 0.555 553 089 945 6;
  • 18) 0.555 553 089 945 6 × 2 = 1 + 0.111 106 179 891 2;
  • 19) 0.111 106 179 891 2 × 2 = 0 + 0.222 212 359 782 4;
  • 20) 0.222 212 359 782 4 × 2 = 0 + 0.444 424 719 564 8;
  • 21) 0.444 424 719 564 8 × 2 = 0 + 0.888 849 439 129 6;
  • 22) 0.888 849 439 129 6 × 2 = 1 + 0.777 698 878 259 2;
  • 23) 0.777 698 878 259 2 × 2 = 1 + 0.555 397 756 518 4;
  • 24) 0.555 397 756 518 4 × 2 = 1 + 0.110 795 513 036 8;
  • 25) 0.110 795 513 036 8 × 2 = 0 + 0.221 591 026 073 6;
  • 26) 0.221 591 026 073 6 × 2 = 0 + 0.443 182 052 147 2;
  • 27) 0.443 182 052 147 2 × 2 = 0 + 0.886 364 104 294 4;
  • 28) 0.886 364 104 294 4 × 2 = 1 + 0.772 728 208 588 8;
  • 29) 0.772 728 208 588 8 × 2 = 1 + 0.545 456 417 177 6;
  • 30) 0.545 456 417 177 6 × 2 = 1 + 0.090 912 834 355 2;
  • 31) 0.090 912 834 355 2 × 2 = 0 + 0.181 825 668 710 4;
  • 32) 0.181 825 668 710 4 × 2 = 0 + 0.363 651 337 420 8;
  • 33) 0.363 651 337 420 8 × 2 = 0 + 0.727 302 674 841 6;
  • 34) 0.727 302 674 841 6 × 2 = 1 + 0.454 605 349 683 2;
  • 35) 0.454 605 349 683 2 × 2 = 0 + 0.909 210 699 366 4;
  • 36) 0.909 210 699 366 4 × 2 = 1 + 0.818 421 398 732 8;
  • 37) 0.818 421 398 732 8 × 2 = 1 + 0.636 842 797 465 6;
  • 38) 0.636 842 797 465 6 × 2 = 1 + 0.273 685 594 931 2;
  • 39) 0.273 685 594 931 2 × 2 = 0 + 0.547 371 189 862 4;
  • 40) 0.547 371 189 862 4 × 2 = 1 + 0.094 742 379 724 8;
  • 41) 0.094 742 379 724 8 × 2 = 0 + 0.189 484 759 449 6;
  • 42) 0.189 484 759 449 6 × 2 = 0 + 0.378 969 518 899 2;
  • 43) 0.378 969 518 899 2 × 2 = 0 + 0.757 939 037 798 4;
  • 44) 0.757 939 037 798 4 × 2 = 1 + 0.515 878 075 596 8;
  • 45) 0.515 878 075 596 8 × 2 = 1 + 0.031 756 151 193 6;
  • 46) 0.031 756 151 193 6 × 2 = 0 + 0.063 512 302 387 2;
  • 47) 0.063 512 302 387 2 × 2 = 0 + 0.127 024 604 774 4;
  • 48) 0.127 024 604 774 4 × 2 = 0 + 0.254 049 209 548 8;
  • 49) 0.254 049 209 548 8 × 2 = 0 + 0.508 098 419 097 6;
  • 50) 0.508 098 419 097 6 × 2 = 1 + 0.016 196 838 195 2;
  • 51) 0.016 196 838 195 2 × 2 = 0 + 0.032 393 676 390 4;
  • 52) 0.032 393 676 390 4 × 2 = 0 + 0.064 787 352 780 8;
  • 53) 0.064 787 352 780 8 × 2 = 0 + 0.129 574 705 561 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:


0.111 111 111 092 3(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0101 1101 0001 1000 0100 0(2)

5. Positive number before normalization:

55.111 111 111 092 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1101 0001 1000 0100 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:


55.111 111 111 092 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1101 0001 1000 0100 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1101 0001 1000 0100 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1110 1000 1100 0010 00(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 1110 1000 1100 0010 00


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 1110 1000 1100 00 1000 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1110 1000 1100


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 1110 1000 1100


Decimal number 55.111 111 111 092 3 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 1110 1000 1100

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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