55.111 111 111 116 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 116 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
55.111 111 111 116 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: 0.111 111 111 116 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) 0.111 111 111 116 3 × 2 = 0 + 0.222 222 222 232 6;
  • 2) 0.222 222 222 232 6 × 2 = 0 + 0.444 444 444 465 2;
  • 3) 0.444 444 444 465 2 × 2 = 0 + 0.888 888 888 930 4;
  • 4) 0.888 888 888 930 4 × 2 = 1 + 0.777 777 777 860 8;
  • 5) 0.777 777 777 860 8 × 2 = 1 + 0.555 555 555 721 6;
  • 6) 0.555 555 555 721 6 × 2 = 1 + 0.111 111 111 443 2;
  • 7) 0.111 111 111 443 2 × 2 = 0 + 0.222 222 222 886 4;
  • 8) 0.222 222 222 886 4 × 2 = 0 + 0.444 444 445 772 8;
  • 9) 0.444 444 445 772 8 × 2 = 0 + 0.888 888 891 545 6;
  • 10) 0.888 888 891 545 6 × 2 = 1 + 0.777 777 783 091 2;
  • 11) 0.777 777 783 091 2 × 2 = 1 + 0.555 555 566 182 4;
  • 12) 0.555 555 566 182 4 × 2 = 1 + 0.111 111 132 364 8;
  • 13) 0.111 111 132 364 8 × 2 = 0 + 0.222 222 264 729 6;
  • 14) 0.222 222 264 729 6 × 2 = 0 + 0.444 444 529 459 2;
  • 15) 0.444 444 529 459 2 × 2 = 0 + 0.888 889 058 918 4;
  • 16) 0.888 889 058 918 4 × 2 = 1 + 0.777 778 117 836 8;
  • 17) 0.777 778 117 836 8 × 2 = 1 + 0.555 556 235 673 6;
  • 18) 0.555 556 235 673 6 × 2 = 1 + 0.111 112 471 347 2;
  • 19) 0.111 112 471 347 2 × 2 = 0 + 0.222 224 942 694 4;
  • 20) 0.222 224 942 694 4 × 2 = 0 + 0.444 449 885 388 8;
  • 21) 0.444 449 885 388 8 × 2 = 0 + 0.888 899 770 777 6;
  • 22) 0.888 899 770 777 6 × 2 = 1 + 0.777 799 541 555 2;
  • 23) 0.777 799 541 555 2 × 2 = 1 + 0.555 599 083 110 4;
  • 24) 0.555 599 083 110 4 × 2 = 1 + 0.111 198 166 220 8;
  • 25) 0.111 198 166 220 8 × 2 = 0 + 0.222 396 332 441 6;
  • 26) 0.222 396 332 441 6 × 2 = 0 + 0.444 792 664 883 2;
  • 27) 0.444 792 664 883 2 × 2 = 0 + 0.889 585 329 766 4;
  • 28) 0.889 585 329 766 4 × 2 = 1 + 0.779 170 659 532 8;
  • 29) 0.779 170 659 532 8 × 2 = 1 + 0.558 341 319 065 6;
  • 30) 0.558 341 319 065 6 × 2 = 1 + 0.116 682 638 131 2;
  • 31) 0.116 682 638 131 2 × 2 = 0 + 0.233 365 276 262 4;
  • 32) 0.233 365 276 262 4 × 2 = 0 + 0.466 730 552 524 8;
  • 33) 0.466 730 552 524 8 × 2 = 0 + 0.933 461 105 049 6;
  • 34) 0.933 461 105 049 6 × 2 = 1 + 0.866 922 210 099 2;
  • 35) 0.866 922 210 099 2 × 2 = 1 + 0.733 844 420 198 4;
  • 36) 0.733 844 420 198 4 × 2 = 1 + 0.467 688 840 396 8;
  • 37) 0.467 688 840 396 8 × 2 = 0 + 0.935 377 680 793 6;
  • 38) 0.935 377 680 793 6 × 2 = 1 + 0.870 755 361 587 2;
  • 39) 0.870 755 361 587 2 × 2 = 1 + 0.741 510 723 174 4;
  • 40) 0.741 510 723 174 4 × 2 = 1 + 0.483 021 446 348 8;
  • 41) 0.483 021 446 348 8 × 2 = 0 + 0.966 042 892 697 6;
  • 42) 0.966 042 892 697 6 × 2 = 1 + 0.932 085 785 395 2;
  • 43) 0.932 085 785 395 2 × 2 = 1 + 0.864 171 570 790 4;
  • 44) 0.864 171 570 790 4 × 2 = 1 + 0.728 343 141 580 8;
  • 45) 0.728 343 141 580 8 × 2 = 1 + 0.456 686 283 161 6;
  • 46) 0.456 686 283 161 6 × 2 = 0 + 0.913 372 566 323 2;
  • 47) 0.913 372 566 323 2 × 2 = 1 + 0.826 745 132 646 4;
  • 48) 0.826 745 132 646 4 × 2 = 1 + 0.653 490 265 292 8;
  • 49) 0.653 490 265 292 8 × 2 = 1 + 0.306 980 530 585 6;
  • 50) 0.306 980 530 585 6 × 2 = 0 + 0.613 961 061 171 2;
  • 51) 0.613 961 061 171 2 × 2 = 1 + 0.227 922 122 342 4;
  • 52) 0.227 922 122 342 4 × 2 = 0 + 0.455 844 244 684 8;
  • 53) 0.455 844 244 684 8 × 2 = 0 + 0.911 688 489 369 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:


0.111 111 111 116 3(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0111 0111 1011 1010 0(2)

5. Positive number before normalization:

55.111 111 111 116 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0111 0111 1011 1010 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:


55.111 111 111 116 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0111 0111 1011 1010 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0111 0111 1011 1010 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 1011 1101 1101 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 0011 1011 1011 1101 1101 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 0011 1011 1011 1101 11 0100 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1011 1011 1101


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 1011 1011 1101


Decimal number 55.111 111 111 116 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 0011 1011 1011 1101

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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