55.111 111 111 062 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 062(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
55.111 111 111 062(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: 0.111 111 111 062.

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) 0.111 111 111 062 × 2 = 0 + 0.222 222 222 124;
  • 2) 0.222 222 222 124 × 2 = 0 + 0.444 444 444 248;
  • 3) 0.444 444 444 248 × 2 = 0 + 0.888 888 888 496;
  • 4) 0.888 888 888 496 × 2 = 1 + 0.777 777 776 992;
  • 5) 0.777 777 776 992 × 2 = 1 + 0.555 555 553 984;
  • 6) 0.555 555 553 984 × 2 = 1 + 0.111 111 107 968;
  • 7) 0.111 111 107 968 × 2 = 0 + 0.222 222 215 936;
  • 8) 0.222 222 215 936 × 2 = 0 + 0.444 444 431 872;
  • 9) 0.444 444 431 872 × 2 = 0 + 0.888 888 863 744;
  • 10) 0.888 888 863 744 × 2 = 1 + 0.777 777 727 488;
  • 11) 0.777 777 727 488 × 2 = 1 + 0.555 555 454 976;
  • 12) 0.555 555 454 976 × 2 = 1 + 0.111 110 909 952;
  • 13) 0.111 110 909 952 × 2 = 0 + 0.222 221 819 904;
  • 14) 0.222 221 819 904 × 2 = 0 + 0.444 443 639 808;
  • 15) 0.444 443 639 808 × 2 = 0 + 0.888 887 279 616;
  • 16) 0.888 887 279 616 × 2 = 1 + 0.777 774 559 232;
  • 17) 0.777 774 559 232 × 2 = 1 + 0.555 549 118 464;
  • 18) 0.555 549 118 464 × 2 = 1 + 0.111 098 236 928;
  • 19) 0.111 098 236 928 × 2 = 0 + 0.222 196 473 856;
  • 20) 0.222 196 473 856 × 2 = 0 + 0.444 392 947 712;
  • 21) 0.444 392 947 712 × 2 = 0 + 0.888 785 895 424;
  • 22) 0.888 785 895 424 × 2 = 1 + 0.777 571 790 848;
  • 23) 0.777 571 790 848 × 2 = 1 + 0.555 143 581 696;
  • 24) 0.555 143 581 696 × 2 = 1 + 0.110 287 163 392;
  • 25) 0.110 287 163 392 × 2 = 0 + 0.220 574 326 784;
  • 26) 0.220 574 326 784 × 2 = 0 + 0.441 148 653 568;
  • 27) 0.441 148 653 568 × 2 = 0 + 0.882 297 307 136;
  • 28) 0.882 297 307 136 × 2 = 1 + 0.764 594 614 272;
  • 29) 0.764 594 614 272 × 2 = 1 + 0.529 189 228 544;
  • 30) 0.529 189 228 544 × 2 = 1 + 0.058 378 457 088;
  • 31) 0.058 378 457 088 × 2 = 0 + 0.116 756 914 176;
  • 32) 0.116 756 914 176 × 2 = 0 + 0.233 513 828 352;
  • 33) 0.233 513 828 352 × 2 = 0 + 0.467 027 656 704;
  • 34) 0.467 027 656 704 × 2 = 0 + 0.934 055 313 408;
  • 35) 0.934 055 313 408 × 2 = 1 + 0.868 110 626 816;
  • 36) 0.868 110 626 816 × 2 = 1 + 0.736 221 253 632;
  • 37) 0.736 221 253 632 × 2 = 1 + 0.472 442 507 264;
  • 38) 0.472 442 507 264 × 2 = 0 + 0.944 885 014 528;
  • 39) 0.944 885 014 528 × 2 = 1 + 0.889 770 029 056;
  • 40) 0.889 770 029 056 × 2 = 1 + 0.779 540 058 112;
  • 41) 0.779 540 058 112 × 2 = 1 + 0.559 080 116 224;
  • 42) 0.559 080 116 224 × 2 = 1 + 0.118 160 232 448;
  • 43) 0.118 160 232 448 × 2 = 0 + 0.236 320 464 896;
  • 44) 0.236 320 464 896 × 2 = 0 + 0.472 640 929 792;
  • 45) 0.472 640 929 792 × 2 = 0 + 0.945 281 859 584;
  • 46) 0.945 281 859 584 × 2 = 1 + 0.890 563 719 168;
  • 47) 0.890 563 719 168 × 2 = 1 + 0.781 127 438 336;
  • 48) 0.781 127 438 336 × 2 = 1 + 0.562 254 876 672;
  • 49) 0.562 254 876 672 × 2 = 1 + 0.124 509 753 344;
  • 50) 0.124 509 753 344 × 2 = 0 + 0.249 019 506 688;
  • 51) 0.249 019 506 688 × 2 = 0 + 0.498 039 013 376;
  • 52) 0.498 039 013 376 × 2 = 0 + 0.996 078 026 752;
  • 53) 0.996 078 026 752 × 2 = 1 + 0.992 156 053 504;

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:


0.111 111 111 062(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0011 1011 1100 0111 1000 1(2)

5. Positive number before normalization:

55.111 111 111 062(10) =

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


55.111 111 111 062(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0011 1011 1100 0111 1000 1(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0011 1011 1100 0111 1000 1(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0001 1101 1110 0011 1100 01(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 0001 1101 1110 0011 1100 01


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 0001 1101 1110 0011 11 0001 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0001 1101 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 0001 1101 1110 0011


Decimal number 55.111 111 111 062 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 0001 1101 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