55.111 111 111 090 7 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 090 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
55.111 111 111 090 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: 0.111 111 111 090 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) 0.111 111 111 090 7 × 2 = 0 + 0.222 222 222 181 4;
  • 2) 0.222 222 222 181 4 × 2 = 0 + 0.444 444 444 362 8;
  • 3) 0.444 444 444 362 8 × 2 = 0 + 0.888 888 888 725 6;
  • 4) 0.888 888 888 725 6 × 2 = 1 + 0.777 777 777 451 2;
  • 5) 0.777 777 777 451 2 × 2 = 1 + 0.555 555 554 902 4;
  • 6) 0.555 555 554 902 4 × 2 = 1 + 0.111 111 109 804 8;
  • 7) 0.111 111 109 804 8 × 2 = 0 + 0.222 222 219 609 6;
  • 8) 0.222 222 219 609 6 × 2 = 0 + 0.444 444 439 219 2;
  • 9) 0.444 444 439 219 2 × 2 = 0 + 0.888 888 878 438 4;
  • 10) 0.888 888 878 438 4 × 2 = 1 + 0.777 777 756 876 8;
  • 11) 0.777 777 756 876 8 × 2 = 1 + 0.555 555 513 753 6;
  • 12) 0.555 555 513 753 6 × 2 = 1 + 0.111 111 027 507 2;
  • 13) 0.111 111 027 507 2 × 2 = 0 + 0.222 222 055 014 4;
  • 14) 0.222 222 055 014 4 × 2 = 0 + 0.444 444 110 028 8;
  • 15) 0.444 444 110 028 8 × 2 = 0 + 0.888 888 220 057 6;
  • 16) 0.888 888 220 057 6 × 2 = 1 + 0.777 776 440 115 2;
  • 17) 0.777 776 440 115 2 × 2 = 1 + 0.555 552 880 230 4;
  • 18) 0.555 552 880 230 4 × 2 = 1 + 0.111 105 760 460 8;
  • 19) 0.111 105 760 460 8 × 2 = 0 + 0.222 211 520 921 6;
  • 20) 0.222 211 520 921 6 × 2 = 0 + 0.444 423 041 843 2;
  • 21) 0.444 423 041 843 2 × 2 = 0 + 0.888 846 083 686 4;
  • 22) 0.888 846 083 686 4 × 2 = 1 + 0.777 692 167 372 8;
  • 23) 0.777 692 167 372 8 × 2 = 1 + 0.555 384 334 745 6;
  • 24) 0.555 384 334 745 6 × 2 = 1 + 0.110 768 669 491 2;
  • 25) 0.110 768 669 491 2 × 2 = 0 + 0.221 537 338 982 4;
  • 26) 0.221 537 338 982 4 × 2 = 0 + 0.443 074 677 964 8;
  • 27) 0.443 074 677 964 8 × 2 = 0 + 0.886 149 355 929 6;
  • 28) 0.886 149 355 929 6 × 2 = 1 + 0.772 298 711 859 2;
  • 29) 0.772 298 711 859 2 × 2 = 1 + 0.544 597 423 718 4;
  • 30) 0.544 597 423 718 4 × 2 = 1 + 0.089 194 847 436 8;
  • 31) 0.089 194 847 436 8 × 2 = 0 + 0.178 389 694 873 6;
  • 32) 0.178 389 694 873 6 × 2 = 0 + 0.356 779 389 747 2;
  • 33) 0.356 779 389 747 2 × 2 = 0 + 0.713 558 779 494 4;
  • 34) 0.713 558 779 494 4 × 2 = 1 + 0.427 117 558 988 8;
  • 35) 0.427 117 558 988 8 × 2 = 0 + 0.854 235 117 977 6;
  • 36) 0.854 235 117 977 6 × 2 = 1 + 0.708 470 235 955 2;
  • 37) 0.708 470 235 955 2 × 2 = 1 + 0.416 940 471 910 4;
  • 38) 0.416 940 471 910 4 × 2 = 0 + 0.833 880 943 820 8;
  • 39) 0.833 880 943 820 8 × 2 = 1 + 0.667 761 887 641 6;
  • 40) 0.667 761 887 641 6 × 2 = 1 + 0.335 523 775 283 2;
  • 41) 0.335 523 775 283 2 × 2 = 0 + 0.671 047 550 566 4;
  • 42) 0.671 047 550 566 4 × 2 = 1 + 0.342 095 101 132 8;
  • 43) 0.342 095 101 132 8 × 2 = 0 + 0.684 190 202 265 6;
  • 44) 0.684 190 202 265 6 × 2 = 1 + 0.368 380 404 531 2;
  • 45) 0.368 380 404 531 2 × 2 = 0 + 0.736 760 809 062 4;
  • 46) 0.736 760 809 062 4 × 2 = 1 + 0.473 521 618 124 8;
  • 47) 0.473 521 618 124 8 × 2 = 0 + 0.947 043 236 249 6;
  • 48) 0.947 043 236 249 6 × 2 = 1 + 0.894 086 472 499 2;
  • 49) 0.894 086 472 499 2 × 2 = 1 + 0.788 172 944 998 4;
  • 50) 0.788 172 944 998 4 × 2 = 1 + 0.576 345 889 996 8;
  • 51) 0.576 345 889 996 8 × 2 = 1 + 0.152 691 779 993 6;
  • 52) 0.152 691 779 993 6 × 2 = 0 + 0.305 383 559 987 2;
  • 53) 0.305 383 559 987 2 × 2 = 0 + 0.610 767 119 974 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:


0.111 111 111 090 7(10) =

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

5. Positive number before normalization:

55.111 111 111 090 7(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0101 1011 0101 0101 1110 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 7(10) =

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

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

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1101 1010 1010 1111 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 1101 1010 1010 1111 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 1101 1010 1010 11 1100 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0010 1101 1010 1010


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 1010 1010


Decimal number 55.111 111 111 090 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 0010 1101 1010 1010

Share

» Convert decimal 55.111 111 111 092 6 to 64 bit double precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 07, 2026 [July]: Decimal Numbers Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: July - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity. NotebookLM YouTube Video of 6.59 minutes

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