I am in 9th grade and taking geometry. Several of my friends taking precalc say that 0.9999... does not equal 1, but is just an asymptote. I have not taken that subject yet and they don't give any proofs for why it is an asymptote. I couldn't find any definate answers on the internet that dealt with asymptotes. It would be much appreciated if someone could explain this to me.

1Asymptote of what? – Sloan May 20 '15 at 16:27

1If $x_0 = 0.9999...$ does not equal $x_1 = 1$, there must be a number $x' \in \mathbb{R}$ such that $x_0 \lt x' \lt x_1$. What would a decimal expansion of such an $x'$ look like ? – collapsar May 20 '15 at 16:30

1Obvious duplicate... – May 20 '15 at 16:33

2As your precalc friends to show you the definition of asymptote (from a book, not just from their muddled memories). Then ask how $0.999\ldots$ fits that definition. – David K May 20 '15 at 17:25

1It would follow from their same logic that $\frac{1}9$ is merely an "asymptote" of $.111\ldots$, since surely no finite sequence of $1$'s is actually $\frac{1}9$. Were that true, it would sort of kills decimal expansion as a reasonable way to represent numbers. – Milo Brandt May 20 '15 at 23:49

There are probably 100 answers on several "$0.999\ldots$ vs $1$" questions on this site. See for example [this](http://math.stackexchange.com/questions/11/isoverline91?lq=1), [this](http://math.stackexchange.com/questions/281492/about09991?lq=1), [this](http://math.stackexchange.com/questions/287311/howcaniexplain0999ldots1?lq=1), [this](http://math.stackexchange.com/questions/567026/whydoes0bar9equal1?rq=1), ... – Winther May 20 '15 at 23:50

If $0.\overline{9}\lt 1.0$ then there exists a positive number $\epsilon$ such that $ \epsilon=1.00.\overline{9}$. Try to show that no such positive number can ever exist (e.g. show that no matter how small it is, it will always be too large). – John Joy May 21 '15 at 13:28
3 Answers
The concept of an asymptote is when some behavior is reached after an infinite amount of steps, which for you would be exactly the idea: any finite number of 9's in the sequence $0.999\ldots$ is indeed strictly less than $1$, but taking an infinite amount of these 9's will make it equal exactly.
It's not easy to define or argue these without formal notions of limits, but here is an intuitive argument. Notice that $$ 0.999\ldots = 9 \times (1/10)^1 + 9 \times (1/10)^2 + 9 \times (1/10)^3 \ldots = 9a (1+a+a^2+\ldots) $$ for $a = 1/10$. If the infinite sum in brackets exists, say it is evaluates to $L$. Then $L*a = a + a^2 + \ldots = L1$, so $aL = L1$ which means that $$ L = \frac{1}{1a} = \frac{1}{11/10} = \frac{10}{9}, $$ and our sum would be $9aL = 9 \times (1/10) \times \frac{10}{9} = 1$, as your intuition suggested.
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Suppose, that $0,99... \ne 1$. In that case, $0,999... < 1$. If $a < b, \exists c$, such that $a < c < b$.
In our case, how this $c$ would look like? It would have the form $0,999...$, but somewhere it can't have $9$. If somewhere we don't write $9$, then $c < a$. Since we can't find such $c$, it is a contradiction. Therefore $0,999...=1$.
Second reason: Consider $\frac13$. It is easy to see, that $\frac13=0,333...$. Multiply both sides with $3$, and you get $1=0,999...$(This is not a proof, just to make it easier to see).
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In what way does this answer the question with respect the users contention that asymptotes are contradicting known proofs that $0.999...=1$? – Sloan May 20 '15 at 16:34

The number 0.99999... is a fixed number, and not the asymptote of anything, even though you might happen to read the number as a succession of values, each corresponding to 1/10th the value of the previous component. I presume you're familiar with the elementary proof that this number equals 1.0, that is, it is just another way of writing the value 1.0:
$10 I = 9.99999...$
$I = 0.99999...$
subtract to find:
$9 I = 9$,
therefore $I = 1.0$.
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