If $0{.}9\ldots$ is $1$, what does that make $0{.}3\ldots$?

Question

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?

Answer 1

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.

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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.

Answer 2

this also proves that two different numbers can be equal.

The point is that they are NOT different numbers, just two different representations.

You've already said that $0{,}\overline{3}$ (The overline is common notation for a repeating set of digits) is $\frac{1}{3}$.

To find out what the other repeating representations are equal to, use the same procedre, i.e. multiply by $10$ (or more generally $10^\text{length of period}$), subtract and divide.

You'll find that $0{,}\overline{x} = \frac{x}{9}$.

Answer 3

Your espression $0.3333...x$ is meaningless.

You can imitate the proof you gave like this: $$ x = 0.333..., \\ 10x = 3.333... \\ 10x-x = 3.000... = 3 \\ 9x = 3 \\ x = 3/9 = 1/3 $$

Your other questions can be answered similarly. $0.7777... = 7/9$ and so on.

Answer 4

It's just that $1$ has $2$ different radix representations in base $10$: $\{1, 0.\bar9\}$. In fact, all integers have $2$ different radix representations in base $10$. The integer itself, and the previous integer $+ \space0.\bar 9$. $13$ can be written as either $13$ or $12.\bar 9$. These two representations point to the same value.

$0.\bar8$ is a radix representation for the rational $\dfrac89$.

You can show this:

$$0.\bar 8 = 8 \left(0.1 + 0.01 + 0.001 + \dots \right) = 8\sum_{k=1}^\infty \left(\frac{1}{10}\right)^k = \frac{\frac{8}{10}}{1 - \frac{1}{10}} = \frac89$$