Convert 0.222 222 222 223 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

0.222 222 222 223(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (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 the binary (base 2) the fractional part: 0.222 222 222 223.

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) 0.222 222 222 223 × 2 = 0 + 0.444 444 444 446;
  • 2) 0.444 444 444 446 × 2 = 0 + 0.888 888 888 892;
  • 3) 0.888 888 888 892 × 2 = 1 + 0.777 777 777 784;
  • 4) 0.777 777 777 784 × 2 = 1 + 0.555 555 555 568;
  • 5) 0.555 555 555 568 × 2 = 1 + 0.111 111 111 136;
  • 6) 0.111 111 111 136 × 2 = 0 + 0.222 222 222 272;
  • 7) 0.222 222 222 272 × 2 = 0 + 0.444 444 444 544;
  • 8) 0.444 444 444 544 × 2 = 0 + 0.888 888 889 088;
  • 9) 0.888 888 889 088 × 2 = 1 + 0.777 777 778 176;
  • 10) 0.777 777 778 176 × 2 = 1 + 0.555 555 556 352;
  • 11) 0.555 555 556 352 × 2 = 1 + 0.111 111 112 704;
  • 12) 0.111 111 112 704 × 2 = 0 + 0.222 222 225 408;
  • 13) 0.222 222 225 408 × 2 = 0 + 0.444 444 450 816;
  • 14) 0.444 444 450 816 × 2 = 0 + 0.888 888 901 632;
  • 15) 0.888 888 901 632 × 2 = 1 + 0.777 777 803 264;
  • 16) 0.777 777 803 264 × 2 = 1 + 0.555 555 606 528;
  • 17) 0.555 555 606 528 × 2 = 1 + 0.111 111 213 056;
  • 18) 0.111 111 213 056 × 2 = 0 + 0.222 222 426 112;
  • 19) 0.222 222 426 112 × 2 = 0 + 0.444 444 852 224;
  • 20) 0.444 444 852 224 × 2 = 0 + 0.888 889 704 448;
  • 21) 0.888 889 704 448 × 2 = 1 + 0.777 779 408 896;
  • 22) 0.777 779 408 896 × 2 = 1 + 0.555 558 817 792;
  • 23) 0.555 558 817 792 × 2 = 1 + 0.111 117 635 584;
  • 24) 0.111 117 635 584 × 2 = 0 + 0.222 235 271 168;
  • 25) 0.222 235 271 168 × 2 = 0 + 0.444 470 542 336;
  • 26) 0.444 470 542 336 × 2 = 0 + 0.888 941 084 672;

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...)


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:


0.222 222 222 223(10) =

0.0011 1000 1110 0011 1000 1110 00(2)


5. Positive number before normalization:

0.222 222 222 223(10) =

0.0011 1000 1110 0011 1000 1110 00(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 3 positions to the right so that only one non zero digit remains to the left of it:


0.222 222 222 223(10) =

0.0011 1000 1110 0011 1000 1110 00(2) =

0.0011 1000 1110 0011 1000 1110 00(2) × 20 =

1.1100 0111 0001 1100 0111 000(2) × 2-3


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): -3

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 =

-3 + 2(8-1) - 1 =

(-3 + 127)(10) =

124(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;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 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) =

124(10) =

0111 1100(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 1100

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


Number 0.222 222 222 223 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 0111 1100 - 110 0011 1000 1110 0011 1000

(32 bits IEEE 754)

More operations of this kind:

0.222 222 222 222 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?

0.222 222 222 224 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

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All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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