What does this highlighted formula means?

Question

Can anyone tell me what the highlighted formula means, how to read this, how to prove it and if possible share a simpler version of this formula?

It seems like it has something to do with infinite periodic decimal fraction. I don't know what it means.

Thanks in advance.

Any help would be appreciated.

Answer 1

I think the author means the following:

Given a rational number $r=p/q$ it has a decimal expansion. And the fact that $r$ is rational is equivalent to the decimal representation being periodic after a certain point. That is if $r=1/3$, then the decimal representation is $r=0.333\cdots$. As another example if you take $r=9/13$, we get $r=0.692307692307\cdots$. It is not necessary that the first few digits after the decimal place should repeat, say for example $r=52/225=0.23111\cdots$. Also for decimal expansions which terminate, we assume that $0$ repeats infinitely, that is if $r=1/2=0.5$, we consider $r=0.5000\cdots$.

In the book, $a$ denotes the integer part of $p/q$, $(\alpha_1,\cdots,\alpha_n)$ denotes the first $n$ non-repeating digits and $(\beta_1\cdots\beta_m)$ denotes repeating digits.

Now given a rational number we can find the corresponding $a$, $\alpha_i$'s and $\beta_j$'s. The reverse representation might mean that the reverse construction is also possible. That is given an integer $a$, $\alpha_i$'s and $\beta_j$ can we find the corresponding $r$.

They have given the formula as $$r=\pm a \pm \frac{\overline{\alpha_1\cdots\alpha_n\beta_1\cdots\beta_m}-\overline{\alpha_1\cdots\alpha_n}}{\overline{\underbrace{99\cdots 9}_{m}\underbrace{0\cdots 0}_{n}}}$$

Here by $\overline{123}$, let's say, I am guessing the author means the number 123 (One hundred and twenty three). I have also made a slight edit to the formula which I think is correct. Hope this helps.

Answer 2

This is the rule of how to convert an infinite periodic decimal fraction into an ordinary fraction.

The result is a fraction where in the numerator one has to subtract a number before the second period and a number before the first period and in the denominator one writes down $m$ nines ($m$ is the number of digits in the period) and $n$ zeros (n is the number of digits before the period). There is a typo in the formula you are providing. The number of nines should be $m$.

For example,

$$ 3{,}28(123) = 3 + 0{,}28(123) = 3 + \frac{28123-28}{99900} = 3 + \frac{28095}{99900} = 3\frac{1873}{6660}. $$

Here,

• $n = 2$:
  • $\alpha_1\alpha_2 = 28$,

• $m = 3$:

  • $\beta_1\beta_2\beta_3 = 123$.

• A number before the second period is 28123.

• A number before the first period is 28.