Why is $\frac{10}{8.1}$ so weird?

Solution 1:

I think you answered your question already:

$$\frac{10}{8.1} = \frac{100}{81} = (1.11111\cdots)^2 = (1.11111\cdots) (1+ 0.1 + 0.01 + 0.001 + \cdots)$$

is really

$$\begin{array}{cr}& 1.1111111111111111 \cdots \\ + &.1111111111111111\cdots \\ + &.0111111111111111\cdots \\ + &.0011111111111111 \cdots \\ \vdots& \vdots \end{array} $$

If you take a look at the $9$-th decimal place, you have nine $1$'s on top, but an extra $1$ coming from the $10$-th decimal place. Then the $9$-th decimal place actually give a $1$ to the $8$-th, therefore killing that $8$ that should be in the $8$-th place.

Solution 2:

actually it is not weird at all.. and you partly solved it $$\frac{10}{8.1} = \frac{100}{81} = \frac{10 \times 10}{9 \times 9}$$ now if you take $999.999.999$ and divide it by $81$ you get $12.345.679$ that's how you can get the period... because now we get $$\frac{100}{81} = 100\frac{12.345.679}{999.999.999}$$ therefore its just constructed to be like this....

Solution 3:

You might actually think of this as having all the numbers 1 to 10 in its period, but since 10 has two digits, you get a carry:

  1.000000000
+ 0.200000000
+ 0.030000000
+ 0.004000000
+ 0.000500000
+ 0.000060000
+ 0.000007000
+ 0.000000800
+ 0.000000090
+ 0.000000010
-------------
carry:    1
-------------
  1.234567900

That this summing is what really happens is explained in Christian Blatters answer, here you see why the 8 is "missing": it get's a carry 1 so it becomes a 9 instead.

Solution 4:

One more remark on this problem: to compute how many repeating decimals a fraction $\frac{p}{q}$ has, one has to compute the order of $10$ modulo $q$ (because then $\left(10^n-1\right)\frac{p}{q} \in \mathbb{Z}$ for the first time).

If we use the "Lifting the exponent lemma", we get that: $$v_3(10^n-1) = v_3(10-1)+v_3(n) = 2+v_3(n) $$ And this should equal $4$ because $81 = 3^4$. So $n = 9$.

Solution 5:

Taking the derivative of ${1\over 1-z}=\sum_{k=0}^\infty z^k$ gives $${1\over(1-z)^2}=\sum_{k=0}^\infty(k+1)z^k\ .$$ If we put $z:={1\over10}$ here we obtain $${10\over 8.1}={1\over\left(1-{1\over10}\right)^2}=\sum_{k=0}^\infty(k+1)\>10^{-k}=1.234567\ldots\quad.$$ (Of course this does not sufficiently explain the later digits.)