A recurring decimal can always be converted into an exact fraction. This is done by using algebra to remove the recurring part of the decimal.

 

Single digit recurring decimals

 

Consider the recurring decimal \( 0.\dot{3} \).

Let
\( x = 0.\dot{3} \)

 

Multiply both sides by \( 10 \) so that the recurring digit lines up:

$$
10x = 3.\dot{3}
$$

 

Now subtract the original equation from this one:

$$
10x - x = 3.\dot{3} - 0.\dot{3}
$$

$$
9x = 3
$$

 

Divide both sides by \( 9 \):

$$
x = \frac{3}{9}
$$

$$
x = \frac{1}{3}
$$

 

So:

$$
0.\dot{3} = \frac{1}{3}
$$

 

 

Two digit recurring decimals

 

Now consider \( 0.\dot{1}\dot{6} \).

Let
\( x = 0.\dot{1}\dot{6} \)

 

Multiply by \( 100 \) because there are two recurring digits:

$$
100x = 16.\dot{1}\dot{6}
$$

 

Subtract the original equation:

$$
100x - x = 16.\dot{1}\dot{6} - 0.\dot{1}\dot{6}
$$

$$
99x = 16
$$

 

Divide both sides by \( 99 \):

$$
x = \frac{16}{99}
$$

 

So:

$$
0.\dot{1}\dot{6} = \frac{16}{99}
$$

 

 

Recurring decimals with a non recurring part

 

Some recurring decimals have digits before the repetition starts. For example, \( 0.2\dot{7} \).

Let
\( x = 0.2\dot{7} \)

 

Multiply by \( 10 \) to move the non recurring digit:

$$
10x = 2.\dot{7}
$$

 

Multiply again by \( 10 \) to cover the recurring digit:

$$
100x = 27.\dot{7}
$$

 

Now subtract:

$$
100x - 10x = 27.\dot{7} - 2.\dot{7}
$$

$$
90x = 25
$$

 

Divide both sides by \( 90 \):

$$
x = \frac{25}{90}
$$

$$
x = \frac{5}{18}
$$

 

So:

$$
0.2\dot{7} = \frac{5}{18}
$$

 

The key idea is to multiply by powers of \( 10 \) so that the recurring digits line up, then subtract to eliminate the repetition. This method works for all recurring decimals and always produces an exact fraction.