0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
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;
  • 0 ÷ 2 = 0 + 0;

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.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118.

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) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 236;
  • 2) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 236 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 472;
  • 3) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 472 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 944;
  • 4) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 944 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 888;
  • 5) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 888 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 776;
  • 6) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 776 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 552;
  • 7) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 552 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 223 104;
  • 8) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 223 104 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 446 208;
  • 9) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 446 208 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 892 416;
  • 10) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 892 416 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 784 832;
  • 11) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 784 832 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 569 664;
  • 12) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 569 664 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 139 328;
  • 13) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 139 328 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 278 656;
  • 14) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 278 656 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 557 312;
  • 15) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 557 312 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 889 114 624;
  • 16) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 889 114 624 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 778 229 248;
  • 17) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 778 229 248 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 556 458 496;
  • 18) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 556 458 496 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 112 916 992;
  • 19) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 112 916 992 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 225 833 984;
  • 20) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 225 833 984 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 451 667 968;
  • 21) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 451 667 968 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 903 335 936;
  • 22) 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 903 335 936 × 2 = 1 + 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 806 671 872;
  • 23) 0.777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 777 806 671 872 × 2 = 1 + 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 613 343 744;
  • 24) 0.555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 555 613 343 744 × 2 = 1 + 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 226 687 488;
  • 25) 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 226 687 488 × 2 = 0 + 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 453 374 976;
  • 26) 0.222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 222 453 374 976 × 2 = 0 + 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 906 749 952;
  • 27) 0.444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 444 906 749 952 × 2 = 0 + 0.888 888 888 888 888 888 888 888 888 888 888 888 888 888 888 889 813 499 904;

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:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

5. Positive number before normalization:

0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118(10) =

0.0001 1100 0111 0001 1100 0111 000(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 4 positions to the right, so that only one non zero digit remains to the left of it:


0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118(10) =

0.0001 1100 0111 0001 1100 0111 000(2) =

0.0001 1100 0111 0001 1100 0111 000(2) × 20 =

1.1100 0111 0001 1100 0111 000(2) × 2-4


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -4

Mantissa (not normalized):
1.1100 0111 0001 1100 0111 000


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-4 + 2(8-1) - 1 =

(-4 + 127)(10) =

123(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 123 ÷ 2 = 61 + 1;
  • 61 ÷ 2 = 30 + 1;
  • 30 ÷ 2 = 15 + 0;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 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) =

123(10) =

0111 1011(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 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 110 0011 1000 1110 0011 1000 =

110 0011 1000 1110 0011 1000


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1011

Mantissa (23 bits) =
110 0011 1000 1110 0011 1000


Decimal number 0.111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 111 118 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1011 - 110 0011 1000 1110 0011 1000

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How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single 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 base ten 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 of the previous dividing 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 previous multiplying operations, starting from the top of the constructed list 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, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits 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 -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 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:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. 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) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) 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:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111