Why do I get $0.098765432098765432...$ when I divide $8$ by $81$?

I got this remarkable thing when I divided $16$ by $162$, or, in a simplified version, $8$ by $81$. It's $0.098765432098765432\cdots$, or more commonly known as $0.\overline{098765432}$, with all the one-digit numbers going backwards...except for $1$. Yeah, it's missing the $1$. One, how do I get this remarkable outcome and two, why is it missing the $1$?

Solution 1:

The link that @Micah posted in the comment to your post is a good explanation.

Maybe you can also see it by the following calculations

$$x = 0.\overline{098765432}$$

$$10^9 x = 98765432.\overline{098765432}$$

$$(10^9 - 1)x = 98765432$$

$$x = 8/81$$

We could actually find the number $y = 0.\overline{0987654321}$ by similar logic.

$$10^{10} y = 987654321.\overline{0987654321}$$

$$(10^{10} - 1) y = 987654321$$

$$y = \frac{109739369}{1111111111}$$

So maybe the reason why you're number is cool is that it has a easy fractional representation and a decimal representation with interesting properties. Any interesting repeating decimal representation is of course a fraction and can be calculated, but it may not have an easy fractional representation as shown with $y$.

Solution 2:

More generally, in base $b$ (any integer $>1$)

$$ \dfrac{b-1}{b^2} + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} = \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}$$ so that $$ \eqalign{ \dfrac{b-1}{b^2}& + \dfrac{b-2}{b^3} + \ldots + \dfrac{2}{b^{b-1}} + \dfrac{0}{b^{b}} + \dfrac{b-1}{b^{b+1}} + \dfrac{b-2}{b^{b+3}} + \ldots = \sum_{j=0}^\infty b^{-j(b-1)} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}\cr &= \dfrac{b^{b-1}}{b^{b-1}-1} \dfrac{(b-2)(b^{b-1} - 1)}{b^{b-1} (b-1)^2}= \dfrac{(b-2)}{(b-1)^2}}$$

What you have is the case $b=10$.