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I recently learned how to convert a repeating decimal like $3.424242...$ to a fraction. I was however wondering why that actually works. After reading a few resources, I tried to understand it from a previous question on this site.

The explanation goes like this:

Let $x=y.a_1a_2\ldots a_m b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$, where $y\in \mathbb N$.

Then $10^m x=t+f$, where $t\in \mathbb N$ and $f=0.b_1b_2\ldots b_p b_1b_2\ldots b_p \ldots$ .

That makes sense so far for me. The following part is where I get lost:

Now, $10^p f=b+f$, where $b=(b_1b_2\ldots b_p)_{10}\in \mathbb N$. So, $f=\dfrac{b}{10^p-1}$

Thus $x=\dfrac{t+\dfrac{b}{10^p-1}}{10^m}=\dfrac{t(10^p-1)+b}{10^m(10^p-1)}$ is a quotient of two natural numbers.

Why do we talk about $10^p f=b+f$ now instead of the $10^m x=t+f$ from the previous lines?

BMBM
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  • This trick only works if the decimal part is fully repetitive, if you get what I mean. The first part only gets rid of the $a_1 ... a_m$. Then you can apply the trick on the $b_1...b_p$ sequence. – TCiur Aug 24 '17 at 07:35
  • $m$ is the length of the non-repeating bit and $p$ is the length of the repeating bit. – Mark Bennet Aug 24 '17 at 07:39

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In the equation $$10^m x = t + f$$ we have two "annoying" variables, $f$ and $t$ (annoying in the sense that they are not written as a fraction). The two variables are $f$ and $x$.

Our goal is to transform $x$ into a fraction, and since $t$ is already a fraction, the equation shows that we will be able to write $x$ as a fraction if we can figure out a way to write $f$ as a fraction. So, our next goal is to discover an equation for $f$ that involves only $f$ and whole numbers.

Which is exactly what the equation $$10^p f = b+f$$ is.

By the way, there is another way of looking at the transformation, and that involves knowing that

$$x=y.a_1a_2\dots a_mb_1b_2\dots b_p b_1b_2\dots b_p\dots =y+\frac{a_1a_2\dots a_m}{10^m} + \frac{b_1}{10^{m+1}} + \frac{b_2}{10^{m+2}} + \cdots$$

You can reduce this to

$$x=y+a' + \frac{1}{10^{m+1}}\left(\frac{b_1}{10}+\frac{b_2}{10^2} + \cdots + \frac{b_p}{10^p} + \frac{b_1}{10^{p+1}}+\dots+\frac{b_p}{10^{p+p}} + \dots\right)$$

where $a'=\frac{a_1a_2\dots a_m}{10^m}$. Now define $b=\frac{b_1}{10}+\frac{b_2}{10^2}+\cdots+\frac{b_p}{10^p}$ and you get

$$x=y+a'+\frac{1}{10^{m+1}}\left(b + \frac{1}{10^p} b + \frac{1}{10^{2p} }b + \cdots\right) \\x=y+a'+\frac{b}{10^{m+1}}\sum_{k=0}^\infty\left(\frac{1}{10^p}\right)^k$$

and using the fact that $\sum_{k=0}^\infty r^k = \frac{1}{1-r}$ you get

$$x=y+a'+\frac{b}{10^{m+1}}\frac{1}{1-10^p}$$

All the numbers in the expression above are fractions, so the result is also a fraction.

5xum
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