Recurring decimals as fractions

Earlier I solved this question and it got me thinking.

\[\text{Why do recurring decimals have a fractional equivalent?}\]

So I decided to work it out, this is what I got.

Let $b$ be a number with $n$ digits, it can have both leading and trailing zeros.

Every recurring decimal can now be represented by the decimal : $0.bbb\ldots$

To represent this as a fraction we need to see it as an infinite sum.

$0.bbb\ldots = \frac{b}{10^{n}} + \frac{b}{10^{2n}} + \frac{b}{10^{3n}} + \ldots$

Now that we have it as an infinite sum we can use the geometric series formula where

$a = \frac{b}{10^{n}}, r = \frac{1}{10^{n}}$

$\sum\limits_{x = 0}^{\infty} ar^{x} = \frac{a}{1 - r}$

Hence

$\sum\limits_{x = 0}^{\infty}\frac{b}{10^{n(x + 1)}} = \frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}}$

$\large\frac{\frac{b}{10^{n}}}{1 - \frac{1}{10^{n}}} = \frac{\frac{b}{10^{n}}}{\frac{10^{n} - 1}{10^{n}}} = \frac{b}{10^{n}} \cdot \frac{10^{n}}{10^{n} - 1} = \frac{b}{10^{n} - 1}$

So that means that :

$0.bbb \ldots = \frac{b}{10^{n} - 1}$

But what if there's some decimal places in front of the recurring part?

Let $c$ be a number with $n_1$ digits, it can also have leading and trailing zeros.

The decimal is now : $0.cbbb \ldots$

Now we have to account for the decimal places between the decimal point and the recurring decimal. This is easy to do since we already have the equation for recurring decimals. The solution is simple - divide by $10$ to the power of the number of places ($n_1$).

$0.cbbb\ldots = \text{ ?} + \frac{b}{10^{n_1}(10^{n} - 1)}$

The $\text{?}$ represents the $c$ part of the decimal. It is simply the number $c$ divided by $10$ to the power of the number of decimal places ($n_1$) it takes up.

$0.cbbb\ldots = \frac{c}{10^{n_1}} + \frac{b}{10^{n_1}(10^{n} - 1)}$

Let's simplify it a bit.

$0.cbbb\ldots = \frac{b + c(10^{n} - 1)}{10^{n_1}(10^{n} - 1)}$

That's all for now, hope you found this note interesting.