Unable to think about an argument in proving that decimal expansion of a rational number m/n is eventually repeating

Question

I could not think about how to prove that decimal expansion of a rational number of a rational number is eventually repeating.

Note that this question has to be proved using Pigeonhole principle only. So, don't close it.

While searching on internet I found this solution whose image I am posting.

But I am unable to think about how does author writes this in third line of proof ->To compute the 10 place digit $a_1$ one computes $r_0$ ×10 = $a_1$ × n +$r_1$ .

Can someone please give explanation of this statement that why this must be true.

Answer 1

If you are struggling to understand some abstract statement first consider a concrete example.

For instance, set $m=17,n=7$. Then $\frac{m}{n}=\frac{17}{7}=2.428571428571...$.

So by division algorithm, you get $17=2\cdot7+3$. You can observe $2$ is the integer part and $3/7$ is the fractional part which is equal to $0.42857142857... $

what is the tenth place digit it is $4$ right? what is it? it is the integer part of $10* 0.42857142857... $ or in other words it is the integer part of $\frac{3\cdot10}{7}$

To get the integer part you multiply $3$ by $10$ an apply division algorithm. which gives you $30=4\cdot7+2$. You can proceed in this manner to get any $10^{-i}$th place digit.

Edit: Suppose $r_i=r_j$ for some $i\lt j$. Also, you know that $0\leq r_i \leq n-1$

Then by division algorithm you can find unique integers $a_{i+1},r_{i+1}$ where $0\leq r_{i+1} \leq n-1$ s.t $10\cdot r_i=a_{i+1}\cdot n+r_{i+1}$.

It follows from the uniqueness that $a_{i+1}=a_{j+1},r_{i+1}=r_{j+1}$

Answer 2

With

$$m = qn + r_0 \tag{1}\label{eq1A}$$

you get

$$\frac{m}{n} = q + \frac{r_0}{n} \tag{2}\label{eq2A}$$

As stated, $q$ is the integer part. Also, $\frac{r_0}{n}$ is the fractional part. In base $10$, if $a_1$ is the first decimal digit, it would be

$$\frac{r_0}{n} = 0.a_1... = \frac{a_1}{10} + r_2 \tag{3}\label{eq3A}$$

where "$\ldots$" are the following digits and

$$0 \le r_2 \lt 0.1 \tag{4}\label{eq4A}$$

To get the $a_1$ as the integer part of an expression, you can multiply both sides of \eqref{eq3A} by $10n$ to get

$$10(r_0) = a_1 n + 10n(r_2) \tag{5}\label{eq5A}$$

You have here that $r_1 = 10n(r_2)$. From \eqref{eq4A}, you get $0 \le r_1 \lt 0.1(10n) = n$. Since $r_1$ must be an integer, this becomes $0 \le r_1 \le n - 1$, as stated in the image.

Answer 3

By division theorem there are $m = qn + r_0$ where $0\le r_0 < n$.

So $\frac mn = q + \frac {r_0}n$.

Now consider the number the number $K = 10r_0$. As $0 \le r_0 < n$ then $0 \le 10r_0 = K < 10n$.

So be division theorem there are $a_1$ and $r_1$ where $K= a_1n + r_1$ where $0 \le r_1 < n$. But notice that $K < 10n$ so $\frac Kn < 10$ so $a_1 < n$ so $a_1 = 0......9$.

So $K = 10r_0 = a_1n + r_1$.

And $m = qn + r_0 = qn + \frac {10r_0}{10} = qn + \frac {a_1n + r_1}{10}$.

So $\frac mn = q +\frac {a_1 + \frac {r_1}n}{10}=$

$q + \frac {a_1}{10} + \frac {r_1}{10n}$.

And as $q_1 = 0......9$ then $\frac {a_1}{10} = 0.a_1$

and $\frac mn = q.a_1 + \frac {r_1}{10n}$.

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A concrete example:

Consider $\frac mn = \frac{596}{73}$ (two number off the top of why head).

Now we want $596= q*73 + r_0$ and we get that by

$m = 596 = 8*73 + 12$ where $12 = r_0$ is the remainder.

So $\frac mn = \frac{596}{73} = 8 + \frac {12}{73}$.

Well, there's nothing in the world to stop us from wondering what $10r_0 = 120$ is like. It might not be clear why we wonder but ... we can wonder.

We want $10r_0 = 120 = 73a_1 + r_1$.

What that is solved by

$530 = 73*1 + 47$ where $a_1 = 1$ and $r_1 = 47$.

But what does that mean?

$\frac mn = 8 + \frac{12}{73}=8 + \frac {120}{73*10}$

$=8 + \frac {73*1+47}{73*10}= 8+\frac {73*1}{73*10} +\frac {47}{73*10}$

$=8 + \frac 1{10} + \frac {47}{73*10}=8.1 + \frac {47}{73*10}$.

And that $1 = a_1$ serves as our first decimal.