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Now this question might sound a bit weird to some people, but the situation is this:

Say I have the number $0.999..$ where there is an infinite number of 9's (much like $0.3333..$ with $\frac{1}{3}$). For convenience, let's put a line above the 9's to indicate there is an infinite number of 9's: $0.\overline{99}$ (probably not official notation, but what the heck..).

Now if I see a statement like this $1=0.\overline{99}$, I would think this is not true, at first, through more or less the same logic that $1\neq2$.

But the following confuses me: $\frac{1}{9}=0.\overline{11}$, so If I do this: $$ \frac{1}{9}\cdot9=\frac{9}{9}=1 $$ And also: $$ 0.\overline{11}\cdot9=0.\overline{99} $$

That means that $1=0.\overline{99}$, but that doesn't make sense to me, because $1$ and $0.\overline{99}$ look like totally different numbers to me... Am I missing something here or is my above logic just false?

Any clarification would be great :)

Mikhail Katz
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Krienie
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3 Answers3

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Your logic is correct.

Consider: $$ \begin{align} x = 0&.9999\ldots\\ 10x = 9&.9999\ldots\\ 9x = 9&\\ x = 1& \end{align} $$

There is an entire Wikipedia entry on this with a slew of proofs of various kinds.

Avraham
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Your logic is correct. $0.999...$ is just a different way of writing the number $1$.

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Here is one way that might convince you of its truth:
Let $x=0.999...$
Then $10x=9.999...=9+0.999...=9+x$
Therefore $10x-x=9\implies x=1$

Mufasa
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  • Another thing to consider is this: "What number could you possibly add to $0.999...$ to get $1$?" – Mufasa Sep 11 '13 at 21:48