55.111 111 111 096 3 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 55.111 111 111 096 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
55.111 111 111 096 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: 0.111 111 111 096 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) 0.111 111 111 096 3 × 2 = 0 + 0.222 222 222 192 6;
  • 2) 0.222 222 222 192 6 × 2 = 0 + 0.444 444 444 385 2;
  • 3) 0.444 444 444 385 2 × 2 = 0 + 0.888 888 888 770 4;
  • 4) 0.888 888 888 770 4 × 2 = 1 + 0.777 777 777 540 8;
  • 5) 0.777 777 777 540 8 × 2 = 1 + 0.555 555 555 081 6;
  • 6) 0.555 555 555 081 6 × 2 = 1 + 0.111 111 110 163 2;
  • 7) 0.111 111 110 163 2 × 2 = 0 + 0.222 222 220 326 4;
  • 8) 0.222 222 220 326 4 × 2 = 0 + 0.444 444 440 652 8;
  • 9) 0.444 444 440 652 8 × 2 = 0 + 0.888 888 881 305 6;
  • 10) 0.888 888 881 305 6 × 2 = 1 + 0.777 777 762 611 2;
  • 11) 0.777 777 762 611 2 × 2 = 1 + 0.555 555 525 222 4;
  • 12) 0.555 555 525 222 4 × 2 = 1 + 0.111 111 050 444 8;
  • 13) 0.111 111 050 444 8 × 2 = 0 + 0.222 222 100 889 6;
  • 14) 0.222 222 100 889 6 × 2 = 0 + 0.444 444 201 779 2;
  • 15) 0.444 444 201 779 2 × 2 = 0 + 0.888 888 403 558 4;
  • 16) 0.888 888 403 558 4 × 2 = 1 + 0.777 776 807 116 8;
  • 17) 0.777 776 807 116 8 × 2 = 1 + 0.555 553 614 233 6;
  • 18) 0.555 553 614 233 6 × 2 = 1 + 0.111 107 228 467 2;
  • 19) 0.111 107 228 467 2 × 2 = 0 + 0.222 214 456 934 4;
  • 20) 0.222 214 456 934 4 × 2 = 0 + 0.444 428 913 868 8;
  • 21) 0.444 428 913 868 8 × 2 = 0 + 0.888 857 827 737 6;
  • 22) 0.888 857 827 737 6 × 2 = 1 + 0.777 715 655 475 2;
  • 23) 0.777 715 655 475 2 × 2 = 1 + 0.555 431 310 950 4;
  • 24) 0.555 431 310 950 4 × 2 = 1 + 0.110 862 621 900 8;
  • 25) 0.110 862 621 900 8 × 2 = 0 + 0.221 725 243 801 6;
  • 26) 0.221 725 243 801 6 × 2 = 0 + 0.443 450 487 603 2;
  • 27) 0.443 450 487 603 2 × 2 = 0 + 0.886 900 975 206 4;
  • 28) 0.886 900 975 206 4 × 2 = 1 + 0.773 801 950 412 8;
  • 29) 0.773 801 950 412 8 × 2 = 1 + 0.547 603 900 825 6;
  • 30) 0.547 603 900 825 6 × 2 = 1 + 0.095 207 801 651 2;
  • 31) 0.095 207 801 651 2 × 2 = 0 + 0.190 415 603 302 4;
  • 32) 0.190 415 603 302 4 × 2 = 0 + 0.380 831 206 604 8;
  • 33) 0.380 831 206 604 8 × 2 = 0 + 0.761 662 413 209 6;
  • 34) 0.761 662 413 209 6 × 2 = 1 + 0.523 324 826 419 2;
  • 35) 0.523 324 826 419 2 × 2 = 1 + 0.046 649 652 838 4;
  • 36) 0.046 649 652 838 4 × 2 = 0 + 0.093 299 305 676 8;
  • 37) 0.093 299 305 676 8 × 2 = 0 + 0.186 598 611 353 6;
  • 38) 0.186 598 611 353 6 × 2 = 0 + 0.373 197 222 707 2;
  • 39) 0.373 197 222 707 2 × 2 = 0 + 0.746 394 445 414 4;
  • 40) 0.746 394 445 414 4 × 2 = 1 + 0.492 788 890 828 8;
  • 41) 0.492 788 890 828 8 × 2 = 0 + 0.985 577 781 657 6;
  • 42) 0.985 577 781 657 6 × 2 = 1 + 0.971 155 563 315 2;
  • 43) 0.971 155 563 315 2 × 2 = 1 + 0.942 311 126 630 4;
  • 44) 0.942 311 126 630 4 × 2 = 1 + 0.884 622 253 260 8;
  • 45) 0.884 622 253 260 8 × 2 = 1 + 0.769 244 506 521 6;
  • 46) 0.769 244 506 521 6 × 2 = 1 + 0.538 489 013 043 2;
  • 47) 0.538 489 013 043 2 × 2 = 1 + 0.076 978 026 086 4;
  • 48) 0.076 978 026 086 4 × 2 = 0 + 0.153 956 052 172 8;
  • 49) 0.153 956 052 172 8 × 2 = 0 + 0.307 912 104 345 6;
  • 50) 0.307 912 104 345 6 × 2 = 0 + 0.615 824 208 691 2;
  • 51) 0.615 824 208 691 2 × 2 = 1 + 0.231 648 417 382 4;
  • 52) 0.231 648 417 382 4 × 2 = 0 + 0.463 296 834 764 8;
  • 53) 0.463 296 834 764 8 × 2 = 0 + 0.926 593 669 529 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:


0.111 111 111 096 3(10) =

0.0001 1100 0111 0001 1100 0111 0001 1100 0110 0001 0111 1110 0010 0(2)

5. Positive number before normalization:

55.111 111 111 096 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 0001 0111 1110 0010 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:


55.111 111 111 096 3(10) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 0001 0111 1110 0010 0(2) =

11 0111.0001 1100 0111 0001 1100 0111 0001 1100 0110 0001 0111 1110 0010 0(2) × 20 =

1.1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0000 1011 1111 0001 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 0000 1011 1111 0001 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 0000 1011 1111 00 0100 =

1011 1000 1110 0011 1000 1110 0011 1000 1110 0011 0000 1011 1111


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 0000 1011 1111


Decimal number 55.111 111 111 096 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 0000 1011 1111

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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