So I recently learned that $0{.}9$ repeating is equal to $1$:
$$ x = 0{.}9\ldots\\ 10x = 9{.}9\ldots\\ 9x = 10x - x = 9{.}9\ldots - 0{.}9\ldots = 9\\ x = 9x/9 = 9/9 = 1\\ x = 1 $$
Or a simpler proof:
$$ x = 1/3 = 0{.}3\ldots\\ 3x = 3/3 = 0{.}9\ldots $$
Which I had no trouble stomaching, this also proves that two different numbers can be equal.
Here's my difficulty though, if the repeating decimal $0{.}9\ldots$ is equal to another, what is $0{.}1\ldots$, $0{.}2\ldots$, $0{.}3\ldots$ etc. up until $0{.}8\ldots$ equal to?
The idea is the same, just solve for $x=0{.}\overline{8}$: $$10x=8{.}\overline{8}=8+x\implies 9x=8\implies x=\frac 89$$
Surely you can do the other fractions on your own, or just note that $0{.}\overline{1}=\frac19$ by the same argument, and therefore $0{.}\overline{2}=2\cdot\frac19=\frac29$ and so on.
Note that the "simpler" proof relies on the fact that $\frac13=0{.}\overline{3}$. If you don't understand why this is the case, then the proof is not simpler, since it sweeps the difficulty under rug, rather than dissolving it.