Finding the repeating decimal form of a fraction $\frac{1}{p}$

Question

I was reading the book Dead Reckoning: Calculating Without Instruments by Ronald W. Doerfler and in it, he mentions a method to convert fractions of the form $\frac{1}{p}, p \in \mathbb{N}$ to a repeating decimal (he uses $t$ instead of $p$). I got what he did there, as in I can now imitate the method and perform the computation but can anyone please explain to me why this method works? I am unable to figure it out myself.

This is the first part of the explanation.

This is the continuation of the same.

I will add that I got through high school recently and I am not very familiar with number theory but is there a way to prove why this works? Thanks in advance!

Edit: I am quoting the text of the author here as well so that it can be read even without accessing the images.

We now turn to the task of finding the decimal expansion of a reciprocal $\frac{1}{t}$. The process of long division, which we can end when a remainder is repeated in the course of a calculation, is the obvious approach. There is another method which is frequently more convenient, particularly if the period of the recurring group is large. We divide by long division until some digits are determined and a low remainder is found. Then we can multiply the entire quotient (including the fraction remainder/$t$) by the remainder, converting the fractional part to proper form. Any digits to the left of the decimal point are discarded. This process of multiplication can be repeated until the recurring group is found, with checks provided by our earlier observations. For example, let us consider the reciprocal of 43: $$\frac{1}{43} = 0.023\frac{11}{43}$$ $$\frac{11}{43}=11 \left(0.023\frac{11}{43} \right) = 0.253 \frac{121}{43} = 0.255 \frac{35}{43}$$ $$\frac{11^2}{43}=11 \left(0.255\frac{35}{43} \right) = 2.805 \frac{385}{43} = 2.813 \frac{41}{43}$$ etc. $$ $$ Then $\frac{1}{43} = 0.023255813 \dots$ $$ $$ We haven't reached the end of a recurring group yet, and of course it may take 42 digits. We could continued our initial long division further and found $\frac{1}{43} = 0.0232558 \frac{6}{43}$. On the other hand, we can continue to $\frac{1}{43} = 0.023255813 \frac{41}{43}$, either by long division or even better as the intermediate result from multiplying by elevens in the calculation given above. Here we can take our remainder as -2 (I find that authors of arithmetic methods often don't take arithmetic literally enoough). Then $$\frac{1}{43} = 0.023255814\frac{-2}{43}$$ $$\frac{-2}{43} = -2\left(0.023255814\frac{-2}{43}\right) = -0.046511628 \frac{+4}{43}$$ $$\frac{(-2)^2}{43} = -2\left(0.046511628\frac{4}{43}\right) = 0.093023256 \frac{-8}{43}$$ See the recurrence beginning in the last result, realizing that $6-\frac{8}{43} = 5\frac{35}{43}$? We find then that $$\frac{1}{43} = 0.\overline{023255813953488372093}$$ where the raised line indicates the recurring group.

Answer 1

Two things are important to this method:

• We can represent 1 as $10^x\over{10^x}$
• When we square a number we square its remainder.

So lets look at my comment example:

$1={100\over 100}$ which we can divide by 97 by dividing the numerator by 97 getting ${1{3\over 97}\over 100}$ we then use that the remainder is in a fractional form that's a multiple of what we start with so if $1\over 97$ starts 0.01 then $3\over 97$ starts 0.03 we shift it to the hundredths place because we still have denominator 100. We get 0.0103 now we have denominator $100^2=10000$ and squaring the remainderwe get 9/97 which in theory is just 9*1/97 which we are in the process of calculating we know 9/97 starts 0927 so appending that we get 0.01030927 squared the denominator we now deal with 81/97 which is .81+0.0243+0.000729+0.00002187=.83505087 that we append. Now we deal with the squaring of of the remainder getting 6561/97=67 remainder 62 answer can double the approximate length every time we do this and Euler's totient theorem gives an upper bound on how long the reptend can be.

Answer 2

Okay, I went back and read it over and over trying to explain it and I figured out a few things.

First, there is notational ambiguity in the text. If I write a mixed fraction such as $2\frac{3}{4}$, then it implies $2 + \frac{3}{4} = 2.75$. However, when the author writes $\frac{1}{43} = 0.023 \frac{11}{43}$, it actually means $\frac{1}{43} = 0.023 +\frac{11}{43} \times 10^{-3}$. So I will now write the example mentioned in the question again and multiply the fractional part with the required power of $10$. $$\frac{1}{43} = 0.023 +\frac{11}{43} \times 10^{-3}$$ $$\begin{align} \frac{11}{43} \times 10^{-3} & = (11 \times 10^{-3})\cdot \left(0.023 +\frac{11}{43} \times 10^{-3}\right) \\ &= 0.253 \times 10^{-3} + \frac{121}{43} \times 10^{-6}\\& = 0.255 \times 10^{-3} + \frac{35}{43} \times 10^{-6} \end{align}$$ $$\begin{align} \frac{11^2}{43} \times 10^{-6} & = (11 \times 10^{-3})\cdot \left(0.255 +\frac{35}{43} \times 10^{-6}\right) \\ &= 2.805 \times 10^{-6} + \frac{385}{43} \times 10^{-9}\\& = 2.813 \times 10^{-6} + \frac{41}{43} \times 10^{-9} \end{align}$$ And so on like the author wrote there too.

Now, as for why this works? Here is my reasoning. Take a number $p \in \mathbb{N}$. Its reciprocal is $\frac{1}{p}$. In decimal form, it can be written as $$\frac{1}{p} = a_1 \times 10^{-1} + a_2 \times 10^{-2} + a_3 \times 10^{-3} + \dots + a_n \times 10^{-n} + \frac{r_1}{p} \times 10^{-n}$$ where $a_i$ is the digit at the $i$th decimal place, $r_1$ is the remainder at which we are choosing to stop the long division, and $n$ is the power of $10$ of the last digit we computed before the fraction. Now, the next step is to multiply the remainder by the quotient.

$$\begin{align} \frac{r_1}{p} \times 10^{-n} & = (r_1 \times 10^{-n}) \cdot \left(a_1 \times 10^{-1} + a_2 \times 10^{-2} + a_3 \times 10^{-3} + \dots + a_n \times 10^{-n} + \frac{r_1}{p} \times 10^{-n}\right)\\ & = (r_1 \times 10^{-n}) \cdot (a_1 \times 10^{-1} + a_2 \times 10^{-2} + a_3 \times 10^{-3} + \dots + a_n \times 10^{-n}) + \frac{{r_1}^2}{p} \times 10^{-2n}\\ \end{align}$$ From this, we can calculate the value of the fraction and add it to the decimal expansion. If we continue further, we will find the value of $\frac{{r_1}^2}{p} \times 10^{-2n}$ and so on.

Now for why the author says to convert the fraction to proper form and ignore the digits to the left of the decimal. When we convert to the proper form, we already take care of the digits on the left. For example in the example above, when we had $\frac{121}{43}$, and in the next step we had $2.813$, we ignored that $2$ cause converting the fraction to proper form had already taken care of it and adding it back would have been redundant. Of course, if we did not convert it to proper form then, we would have fractions like $\frac{11^3}{43} = \frac{1331}{42}$ which would be a pain. If we already have the proper form, this situation does not arise and we can continue without worrying about it. In every step, we are only evaluating what the value of the fraction at the end is, and that is why this method works.

So, is this reasoning correct?