Arth from Tanglin Trust School in Singapore considered the case $x = 2.\dot2$:

If $x=0.\dot2$ and we have to get to $2.\dot2$

Then $x+2 = 2.\dot2$

Sanika P from PSBBMS in India continued:

$10x = 2.\dot2$

Therefore,

$\begin{align} x+2&=10x\\

9x&=2\\

x&=\tfrac29\end{align}$

Josh from Winchcombe School in the UK, Mahdi from Mahatma Gandhi International School in India and Sanika P also expressed $0.\dot2\dot5$ as a fraction. This is Sanika's work:

$y = 0.2525...$

$y+25 = 25.2525...$

$y\times100 = 25.2525...$

Therefore,

$\begin{align} y\times100&=y+25\\

99y&=25\\

y&=\tfrac{25}{99}\end{align}$

Jacob from Acland Burghley in the UK, Dylan from Thomas Starr King Middle School in the USA and Sanika P all described a general method for writing decimals like this as fractions. Dylan wrote:

Let’s call this decimal $x.$ If all of $x$ repeats, like $0.\dot1\dot5,$ then multiply $x$ by a multiple [power] of $10$ so that $1$ repetition of $x$ is on the left of the decimal point. Now subtract $x$ from that value and now $x$ should have a coefficient equal to a whole number. Divide the whole number by the coefficient of $x$ and you have a fraction.

Jacob wrote a method which almost jumps straight to the answer:

1. Multiply $n$ by $10^\text{the number of recurring digits}$

2. Subtract $n$ to get $n \times (10^\text{the number or recurring digits} - 1)$

3. Divide by $10^\text{the number or recurring digits} - 1$

4. Done

And Dylan added a short cut to jump straight to the answer:

This all can be shortcut by taking one repetition and putting it over a number made of $9$s with the same [number] of digits, I.e. $0.\dot3\dot2$ equals $\frac{32}{99}.$ Of course if you need to explain how you got your answer then the method above shows each step.

*See the problem Tiny Nines for more on this idea*

Josh wrote this using sigma notation. Click to see Josh's work.

A recurring decimal $z$ that has $n$* *alternating digits and can be represented in the form$$z=\sum_{i=0}^{\infty}{\frac{1}{{10}^{in}}\left(\sum_{j=1}^{n}\frac{a_j}{{10}^j}\right)}$$ where $a_j$ are integers between $1$ and $9$ inclusive that represent the [recurring] digits.

Then $$z=\frac{{10}^n}{{10}^n-1}\sum_{j=1}^{n}\frac{a_j}{{10}^j}=\frac{1}{{10}^n-1}\sum_{j=1}^{n}{{10}^{n-j}a_j}$$

Then $$z=\frac{{10}^n}{{10}^n-1}\sum_{j=1}^{n}\frac{a_j}{{10}^j}=\frac{1}{{10}^n-1}\sum_{j=1}^{n}{{10}^{n-j}a_j}$$

Sanika P and Mahdi also expressed the other recurring decimals as fractions. This is Mahdi's work:

Luke, Dylan and Sanika P all described a general method for recuring decimals like these. This is Sanika's method:

Let us take the decimal $0.23444...$ to elaborate.

Let $x$ be equal to $0.2344...$

We first multiply $x$ by $100$ to get a fully recurrent decimal part.

$x\times100= 23.44...$

We then multiply $x$ by $1000$ so that exactly one 1 iteration comes to the whole number part: $x\times1000=234.444...$

Now, we have obtained the same decimal part for both the case which makes it easier for us to subtract [one equation from the other] to get the following:

$x\times900=211$

Therefore, $x= \frac{211}{900}.$

Sanika also used a slightly different method for one of the examples:

$y = 0.8333...$

$\begin{align}8.333... &= y\times10\\ 8.333 &=y+7.5\end{align}$

Therefore,

$\begin{align}y\times10&=y+7.5\\ 9y&=7.5 \\ y&=\tfrac{75}{90}\end{align}$

Dylan described this method:

If not all of the decimal repeats, like $0.8\dot3,$ then the process is slightly different. Let us still call this decimal $x.$ First you need to multiply $x$ by a multiple [power] of $10$ so that the part that does not repeat is on the left of the decimal point. Now subtract $x$ from this value. You should get a number that has a few decimal points but does
terminate. Let us call this value $y.$

*(Note that you might need to multiply by a different power of $10$ to make sure that the recurring digits line up)*

Now set up an equation with the $x$ multiplied by the [power of $10$] on one side and $x+y$ on the other. Solve this equation to find $x$ as a fraction. You may need to multiply the whole equation by a value equal to $1$ to remove any decimals from the fraction, or simplify.

Mahdi generalised Sanika's method algebraically. Click to see Mahdi's work.

Thông-sahge from Tiffin School in the UK and Mahdi also used another method to convert the recurring decimals into fractions. Thông-sahge also described the process which led to the algebraic generalisation like Mahdi's (above) - linking up the two methods. This is some of Thông-sahge's work:

Method 2: Using $S = \dfrac{a}{1-r}$

*(This is the formula for the sum $S$ of an infinite geometric series, where the first term is $a$ and the common ratio between each term and the next is $r$)*

I know that, for example $0.\dot40\dot5$ is an infinite sum of [the series]:

$0.405 + 0.000405 + 0.000000405 + ...$

Which can be written as:

$\begin{align} S &= \tfrac{0.405}{1-0.001}\\ \Rightarrow S &=\tfrac{405}{999}\end{align}$

Though I could have stopped here, I wondered whether there was an underlying pattern.

So I tried $1.23\dot4\dot5$

$\begin{align} 1.23\dot4\dot5 &=1 + 0.23 + 0.0045 + 0.000045 + ...\\&= 1 + \tfrac{23}{100} + \tfrac{0.0045}{1-0.01}\\&= 1 + \tfrac{23}{100} + \tfrac{45}{9900}\\& = 1 + \tfrac{23\times99+45}{9900}\\& = 1 + \tfrac{23\times(100-1) + 45}{9900}\\&= 1 + \tfrac{2345 - 23}{9900}\end{align}$

I tried more examples, and I then noticed, in my example,

$\begin{align} 1 + \tfrac{2345 - 23}{9900}&= \tfrac{1(9900) + 2345 - 23}{9900}\\& = \tfrac{12345 - 123}{9900}\end{align}$

Wow!!! So if I generalise: $$c_1c_2...c_lb_1b_2...b_m\dot{a_1}a_2...\dot{a_n} = \frac{\left(c_1c_2...c_l.b_1b_2...b_ma_1a_2...a_n\right) - \left(c_1c_2...c_lb_1b_2...b_m\right)}{\underbrace{999...9}_{n\text{ 9s}}\underbrace{000...0}_{m\text{ 0s}}}$$ (for some [positive integers] $l,m,n$)

Which is the same as Mahdi's result.