And yet another chance to brag about my $$\LaTeX$$!

[size=200]PART A[/size]
Note: all these numbers are in base three
We can first see that $\frac12$ is equal to $\frac13 + \frac16$
Distributing this further we can see a sequence:
$$\frac12=\frac13+\frac16=\frac13+\frac19+\frac{1}{18}$$
Do you not see the sequence?? Lemme show you more in depth.
$$\frac12=\frac{1}{3^1}+\frac{1}{6}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{18}=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}......$$
Now do you see it?

Well, how will that pattern help us?
$$\frac{1}{3^1}_3\leadsto 0.1$$
$$\frac{1}{3^2}_3\leadsto 0.01$$
$$\frac{1}{3^3}_3\leadsto 0.001$$
We can keep on going but we will just keep on going down adding an extra $0$.
So then we will get a result of $\boxed{\bold{0.\overline{1}}}$



[size=200]PART B[/size]
note:all these numbers are in base 8
We can use the same concept we used in Part A.
$$\frac17=\frac18+\frac{1}{16}$$
As we start to expand, we notice the same pattern!
$$\frac17=\frac18+\frac{1}{16}\leadsto \frac17=\frac18+\frac{1}{64}+\frac{1}{512}.......\leadsto \frac17=0.1+0.01+0.001.....$$

Hmm. The same result! That's fishy

[size=200]PART C[/size]
note:all these numbers are in base $b$
Now let's study $\frac{1}{b-1}$ in base $b$.

To simplify this, let's first make an equation.

$$n=\frac{1}{b-1}$$
We can now easily state that:
$$n=\frac{1}{b}+\frac{1}{b^2}+\frac{1}{b^3}.....$$
$$\rightarrow \cdot b$$
$$bn=1+\frac{1}{b}+\frac{1}{b^2}+.....$$
We can now subtract from out original equation:
$$bn=1+\frac{1}{b}+\frac{1}{b^2}+.....$$
$$-n=-\frac{1}{b}-\frac{1}{b^2}-\frac{1}{b^3}.....$$
$$bn-n=1$$
We can now prove that these two terms are equal
$$bn-n=1$$
$$b(n-1)=1$$

So no matter what, the answer will ALWAYS be $\boxed{\bold{0.\overline{1}}}$