55.111 111 111 111 111 106 19 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 111 111 106 19(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
55.111 111 111 111 111 106 19(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: 0.111 111 111 111 111 106 19.

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) 0.111 111 111 111 111 106 19 × 2 = 0 + 0.222 222 222 222 222 212 38;
  • 2) 0.222 222 222 222 222 212 38 × 2 = 0 + 0.444 444 444 444 444 424 76;
  • 3) 0.444 444 444 444 444 424 76 × 2 = 0 + 0.888 888 888 888 888 849 52;
  • 4) 0.888 888 888 888 888 849 52 × 2 = 1 + 0.777 777 777 777 777 699 04;
  • 5) 0.777 777 777 777 777 699 04 × 2 = 1 + 0.555 555 555 555 555 398 08;
  • 6) 0.555 555 555 555 555 398 08 × 2 = 1 + 0.111 111 111 111 110 796 16;
  • 7) 0.111 111 111 111 110 796 16 × 2 = 0 + 0.222 222 222 222 221 592 32;
  • 8) 0.222 222 222 222 221 592 32 × 2 = 0 + 0.444 444 444 444 443 184 64;
  • 9) 0.444 444 444 444 443 184 64 × 2 = 0 + 0.888 888 888 888 886 369 28;
  • 10) 0.888 888 888 888 886 369 28 × 2 = 1 + 0.777 777 777 777 772 738 56;
  • 11) 0.777 777 777 777 772 738 56 × 2 = 1 + 0.555 555 555 555 545 477 12;
  • 12) 0.555 555 555 555 545 477 12 × 2 = 1 + 0.111 111 111 111 090 954 24;
  • 13) 0.111 111 111 111 090 954 24 × 2 = 0 + 0.222 222 222 222 181 908 48;
  • 14) 0.222 222 222 222 181 908 48 × 2 = 0 + 0.444 444 444 444 363 816 96;
  • 15) 0.444 444 444 444 363 816 96 × 2 = 0 + 0.888 888 888 888 727 633 92;
  • 16) 0.888 888 888 888 727 633 92 × 2 = 1 + 0.777 777 777 777 455 267 84;
  • 17) 0.777 777 777 777 455 267 84 × 2 = 1 + 0.555 555 555 554 910 535 68;
  • 18) 0.555 555 555 554 910 535 68 × 2 = 1 + 0.111 111 111 109 821 071 36;
  • 19) 0.111 111 111 109 821 071 36 × 2 = 0 + 0.222 222 222 219 642 142 72;
  • 20) 0.222 222 222 219 642 142 72 × 2 = 0 + 0.444 444 444 439 284 285 44;
  • 21) 0.444 444 444 439 284 285 44 × 2 = 0 + 0.888 888 888 878 568 570 88;
  • 22) 0.888 888 888 878 568 570 88 × 2 = 1 + 0.777 777 777 757 137 141 76;
  • 23) 0.777 777 777 757 137 141 76 × 2 = 1 + 0.555 555 555 514 274 283 52;
  • 24) 0.555 555 555 514 274 283 52 × 2 = 1 + 0.111 111 111 028 548 567 04;
  • 25) 0.111 111 111 028 548 567 04 × 2 = 0 + 0.222 222 222 057 097 134 08;
  • 26) 0.222 222 222 057 097 134 08 × 2 = 0 + 0.444 444 444 114 194 268 16;
  • 27) 0.444 444 444 114 194 268 16 × 2 = 0 + 0.888 888 888 228 388 536 32;
  • 28) 0.888 888 888 228 388 536 32 × 2 = 1 + 0.777 777 776 456 777 072 64;
  • 29) 0.777 777 776 456 777 072 64 × 2 = 1 + 0.555 555 552 913 554 145 28;
  • 30) 0.555 555 552 913 554 145 28 × 2 = 1 + 0.111 111 105 827 108 290 56;
  • 31) 0.111 111 105 827 108 290 56 × 2 = 0 + 0.222 222 211 654 216 581 12;
  • 32) 0.222 222 211 654 216 581 12 × 2 = 0 + 0.444 444 423 308 433 162 24;
  • 33) 0.444 444 423 308 433 162 24 × 2 = 0 + 0.888 888 846 616 866 324 48;
  • 34) 0.888 888 846 616 866 324 48 × 2 = 1 + 0.777 777 693 233 732 648 96;
  • 35) 0.777 777 693 233 732 648 96 × 2 = 1 + 0.555 555 386 467 465 297 92;
  • 36) 0.555 555 386 467 465 297 92 × 2 = 1 + 0.111 110 772 934 930 595 84;
  • 37) 0.111 110 772 934 930 595 84 × 2 = 0 + 0.222 221 545 869 861 191 68;
  • 38) 0.222 221 545 869 861 191 68 × 2 = 0 + 0.444 443 091 739 722 383 36;
  • 39) 0.444 443 091 739 722 383 36 × 2 = 0 + 0.888 886 183 479 444 766 72;
  • 40) 0.888 886 183 479 444 766 72 × 2 = 1 + 0.777 772 366 958 889 533 44;
  • 41) 0.777 772 366 958 889 533 44 × 2 = 1 + 0.555 544 733 917 779 066 88;
  • 42) 0.555 544 733 917 779 066 88 × 2 = 1 + 0.111 089 467 835 558 133 76;
  • 43) 0.111 089 467 835 558 133 76 × 2 = 0 + 0.222 178 935 671 116 267 52;
  • 44) 0.222 178 935 671 116 267 52 × 2 = 0 + 0.444 357 871 342 232 535 04;
  • 45) 0.444 357 871 342 232 535 04 × 2 = 0 + 0.888 715 742 684 465 070 08;
  • 46) 0.888 715 742 684 465 070 08 × 2 = 1 + 0.777 431 485 368 930 140 16;
  • 47) 0.777 431 485 368 930 140 16 × 2 = 1 + 0.554 862 970 737 860 280 32;
  • 48) 0.554 862 970 737 860 280 32 × 2 = 1 + 0.109 725 941 475 720 560 64;
  • 49) 0.109 725 941 475 720 560 64 × 2 = 0 + 0.219 451 882 951 441 121 28;
  • 50) 0.219 451 882 951 441 121 28 × 2 = 0 + 0.438 903 765 902 882 242 56;
  • 51) 0.438 903 765 902 882 242 56 × 2 = 0 + 0.877 807 531 805 764 485 12;
  • 52) 0.877 807 531 805 764 485 12 × 2 = 1 + 0.755 615 063 611 528 970 24;
  • 53) 0.755 615 063 611 528 970 24 × 2 = 1 + 0.511 230 127 223 057 940 48;

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:


0.111 111 111 111 111 106 19(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2)

5. Positive number before normalization:

55.111 111 111 111 111 106 19(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(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:


55.111 111 111 111 111 106 19(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1100 0111 0001 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 1110 0011 1000 11(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 1000 1110 0011 1000 11


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 1000 1110 0011 10 0011 =

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


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 1000 1110 0011


Decimal number 55.111 111 111 111 111 106 19 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 1000 1110 0011

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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