@V8archie - If you can explain to me what recurring decimals do actually mean then I would be grateful.

You will need to learn about the theory of limits principally. What follows is an imprecise overview of what the theory of limits, what the real numbers are and what decimal notation means.

Essentially, we can imagine infinite sequences of numbers.

Some of these infinite sequences converge to a single value. An easy example is that the sequence $$\frac11, \, \frac12, \, \frac13, \, \frac14, \, \cdots$$ converges to zero. That is, the terms of the sequence get closer and closer to zero - and we can get the sequence arbitrarily close to zero by selecting sufficient terms. One definition of the real numbers is that they are the set of limits of all convergent infinite sequences.

If we add together the terms of an infinite sequence, we get a series - an infinite sum. $$\frac11 + \frac12 + \frac13 + \frac14 + \cdots$$

Now, it's not actually possible to add together an infinite number of terms - if you are using a calculator, you never get to press the "equals" button - but by considering the partial sums of this series, we can create a new sequence$$\frac11, \, \frac11 + \frac12, \, \frac11 + \frac12 + \frac13, \, \cdots$$ and we define the value of the infinite series, or the more loosely the result of the infinite sum, to be the limit of the sequence of these partial sums as we take more and more terms. (As long as such a limit exists).

The decimal representation of numbers is precisely a representation of some of these sequences. So the number $a_0.a_1a_2a_3a_4a_5\ldots$ represents the infinite sum $$a_0 + \frac{a_1}{10} + \frac{a_2}{100} + \frac{a_3}{1000} + \frac{a_4}{10000} + \frac{a_5}{100000} + \cdots$$

More specifically, the number $0.33333\ldots$ represents the infinite sum $$0 + \frac3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \frac3{100000} + \cdots$$

We can show that every single one of these infinite sums (depicted by a decimal representation) converges, and thus for each one the value to which it converges is the value that the decimal representation represents.

In particular, we can say that \begin{align}\text{if} \quad S_n &= 1 + x + x^2 + x^3 + \cdots + x^{n-1} + x^n \\ \text{then} \quad xS_n &= \phantom{1 + {}} x + x^2 + x^3 + \cdots + x^{n-1} + x^n + x^{n+1} \\ \text{and by subtracting the second line from the first} \qquad \\ S_n - xS_n &= 1 + \phantom{x + x^2 + x^3 + \cdots + x^{n-1} + x^n} - x^{n+1} \\ (1-x)S_n &= 1 - x^{n+1} \\ S_n &= \frac{1-x^{n+1}}{1-x}\end{align}

Now, when $-1 \lt x \lt 1$, the value of $x^{n+1}$ gets smaller and smaller as $n$ increases so we can say that the limit of the partial sums $S_n$, which is the infinite sum, is $$1 + x + x^2 + \ldots = \frac1{1-x}$$

Thus, in the case of our $0.333\ldots$ we get \begin{align}0 + \frac3{10} + \frac3{100} + \frac3{1000} + \frac3{10000} + \frac3{100000} + \cdots &= \frac{3}{10}\left(1 + \frac{1}{10} + \left(\frac{1}{10}\right)^2 + \left(\frac{1}{10}\right)^3 + \cdots \right) \\ \text{here we have $x=\frac1{10}$ so} \qquad &= \frac{3}{10}\left(\frac{1}{1-\frac1{10}}\right) \\ &= \frac{3}{10} \left(\frac{1}{\frac{9}{10}}\right) \\ &= \frac{3}{10} \cdot \frac{10}{9} \\ &= \frac39 \\ &= \frac13 \end{align}

You can find an introduction to limits in any beginners book on Calculus.