Consider the sequence $R_n$ of repunits, defined as $\displaystyle\frac{10^n1}{9}$. We have $$\frac{R_{n+1}}{R_n}=\frac{9}{9} \frac{10^{n+1}1}{10^n 1}=\frac{10^{n+1}1}{10^n 1}=\frac{\overbrace{99\cdots99}^{n \ \text{nines}}}{\underbrace{99\cdots99}_{n1 \ \text{nines}}}=\frac{100\cdots00910}{\underbrace{99\cdots9}_{n1 \ \text{nines}}}=10.\overline{00\cdots09},$$ (where the first sequence of zeros contains $n$ of them and the second $n1$), and thus $$\frac{R_2}{R_1}=\frac{99}{9}=11\cdot\frac99=10.\bar9. $$Let $n.d_0d_1d_2\cdots;\cdots d_{H1}d_{H}d_{H+1}\cdots$ be the decimal representation of a hyperreal number between $n$ and $n+1$. To the left of the semicolon one finds the standard decimal part, the nonstandard one to the right of it. Since we do not want $\frac99<1$, isn't the equality above an argument for the representation $n.999\cdots;\cdots999\cdots$, perhaps more succintly expressable as $n.\overset{\infty}{\bar9}$, of $n+1$? Then again, how does that combine with the nonexistence of $0.000\cdots;\cdots999\cdots$ (all the digits after the semicolon are nines)? Do we need to deny the existence of $0.999\cdots;\cdots000\cdots$?
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This seems relevant: http://math.stackexchange.com/questions/281492/about09991 – David K Oct 16 '15 at 01:38

1Indeed, $0.999\cdots;\cdots000\cdots$ with only zeros after the semicolon, is *not* a hyperreal number. – Vincenzo Oliva Mar 19 '17 at 16:34

And, indeed, $0.999\cdots;\cdots999\cdots$ is a thing and equals $1$. – Vincenzo Oliva Mar 19 '17 at 20:07