# How do you find the repeating decimal 0.231 with 31 repeated as a fraction?

Oct 20, 2016

The answer is $\frac{1154}{4995}$

#### Explanation:

$0.2 \left(31\right) = 0.2 + 0.0 \left(31\right) = 0.2 + 0.031 + 0.00031 + 0.0000031 + \ldots$

Starting from the second number we have an infinite sum of a convergent geometric sequence with:

${a}_{1} = 0.031$ and $q = 0.001$

So we can write the whole decimal as:

$0.2 \left(31\right) = 0.2 + \frac{0.031}{1 - 0.001} = 0.2 + \frac{0.031}{0.999} = \frac{2}{10} + \frac{31}{999} =$

$= \frac{2 \cdot 999 + 310}{9990} = \frac{1198 + 310}{9990} = \frac{2308}{9990} = \frac{1154}{4995}$

Oct 20, 2016

$x = \frac{229}{990}$

#### Explanation:

let $x = 0.2313131 \ldots \ldots \ldots \ldots .$

because TWO digits repeat $\times 100$

we have therefore:

$100 x = 23.13131313131313 \ldots \ldots .$.

& x=0.231313131......

Subtracting the two the repeating 31s will all cancel

$100 x = 23.1 \cancel{3131313131313. \ldots \ldots}$.

$x = 0.2 \cancel{31313131. \ldots . .}$

we are left with

$110 x - x = 23.1 - 0.2$

$99 x = 22.9$

$x = \frac{22.9}{99} = \frac{229}{990}$