Is 0.9 repeating = 1 disproved by asymptotes?

Question

I'm discussing proofs that 0.9 repeating equals 1 with some friends, and they use asymptotes to disprove this. One says if we had the function $y=x/0.000\ldots1$ (and he's only using that impossible number for theoretical purposes), the slope of the asymptote would be as close to undefined as you can get, but the value would be 0.9 repeating (if you let 1 represent infinity).

I don't know that much about asymptotes yet, but I'm comparing it to an infinite series, because he says that the function's graph forever approaches the number that is the value of the asymptote, but effectively never reaches it. The infinite series $1/2 + 1/4 + 1/8+\cdots$ forever approaches $1$, and you could say that it effectively never does, but even still the answer is exactly $1$. Is there a better argument against the asymptote argument?

EDIT: I know about the many valid proofs for this, but my problem is that they all refuse to accept those proofs. They're throwing invalid arguments my way, and since they reject mine, I'm stuck having to disprove theirs. I know those arguments can be disproved, just not how, and my question is how to disprove this one. Just clarifying.

Answer 1

$1/2 + 1/4 + 1/8 + \ldots$ is just a number. It doesn't "approach" anything: it simply is $1$.

It is the sequence of partial sums that approaches $1$: that is, the list of numbers

• $1/2 = 1/2$
• $3/4 = 1/2 + 1/4$
• $7/8 = 1/2 + 1/4+ 1/8$
• $15/16 = 1/2 + 1/4+ 1/8 + 1/16$
• ...

is a sequence of numbers whose limit is $1$.

$1/2 + 1/4 + 1/8 + \ldots$ is not this sequence: it is the number that is equal to the limit of this sequence.

Answer 2

No, there are many valid proofs on wikipedia.

Answer 3

"if you let 1 represent infinity" is at best vague language, but I am guessing that it means an infinite number of $0$s precede the digit $1$.

A problem with that is that (at least when one is doing things as they are usually done) every digit after the decimal point in the decimal expansion of a number has only finitely many other digits after the decimal point before it.

One way of viewing the statement that $\displaystyle \frac12+\frac14+\frac18+\frac1{16}+\cdots=1$ is that $1$ is the very smallest number that can never be exceeded if you stop summing after some finite number of terms. Any number even very slightly smaller than $1$ will be passed after some finite number of terms. The situation becomes more involved when one considers series with some positive and some negative terms, but considering only positive terms is enough to account for decimal expansions.

Answer 4

Your friends do not understand what an asymptote is/means "...because he says that the function's graph forever approaches the number that is the value of the asymptote, but effectively never reaches it." He has just admitted that $0.999...9$ ("forever") is equal to $1$ because he admits it "asymptotes" to $1$. This is the same as saying that the limit is equal to $1$: $\lim_{n \rightarrow \infty} \frac{1}{10}\sum_0^n\left(\frac{9}{10}\right)^i = \frac{1}{10}\cdot\frac{1}{1 - \frac{9}{10}} = \frac{1}{10}\frac{1}{\frac{1}{10}} = \frac{1}{10}\frac{10}{1} = 1$. The problem that your friend has is that he has no imagination! It is true that $0.999 \neq 1$, $0.999999 \neq 1$, $0.999999999 \neq 1$, $0.999999999999 \neq 1$, etc. But we are not talking about a finite amount of 9's, we're talking about an infinite amount of 9's!

When someone says that an asymptote is something that a function approaches but never reaches what they literally mean is that it reaches the asymptote in the limit that you go through an infinite number of iterations.

Answer 5

You're headed down the right track with the infinite series argument.

\begin{align} 0.\bar{9} & =\frac{1}{10}\sum_{n=0}^{\infty}9\cdot10^{-n}\\ & =\frac{9}{10}\sum_{n=0}^{\infty}(\frac{1}{10})^{-n}\\ & =\frac{9}{10}\frac{1}{1-\frac{1}{10}}\\ & =\frac{9}{10}\frac{1}{\frac{9}{10}}\\ & =1 \end{align}

The problem with the asymptotic arguments are that they ignore the difference between a minimum and an infimum of sets. In particular, an open interval does not have a minimum or a maximum but may have an infimum or supremum.

Answer 6

A handy, albeit somewhat sloppy way to see $1 = 0.99\ldots$:

Since $0.99\ldots$ has no "last" digit by definition, i.e. since given any $9$ there is a next $9$, so for any real $\varepsilon > 0$ there is an integer $n>0$ such that if $0.99\ldots 9$ has $n$ $9$'s then $$1 - 0.99\ldots 9 < \varepsilon.$$ But then we perforce conclude that $1 = 0.99\ldots$

Answer 7

Study up on Zeno's paradoxes and turn your friends' logic against them.

As for the "function" $ y=x/0.000\ldots1 :$ what is that figure "$0.000\ldots1$" in the denominator? Do your friends claim it is a real number? If it is, then you can divide it by $2,$ or even by $10,$ because those are things you can do to every real number. (In this particular case it's easy to show exactly what the result is, too: just erase the last digit "$1$" and write "$05"$ or "$01$" in its place.) If that thing is not a real number, on the other hand, what kind of function can we define with it? Is that "function" a function at all?

Answer 8

$0.9$ repeating is equivalent to $1$ because the difference of them is $0.0000...1$ or $0.1$ to the power of infinity, but if there is an infinite amount of zeros there's no room for the one. You might say that it will always have enough room for one because it's infinite, but if you can add a one you can add a zero, making the number invalid for the difference, so if there can't be a one at the end it means it can't exist, therefore saying that there is no difference between $0.9$ repeating and $1$ because the difference between them cannot exist.

Answer 9

$$x=0.999...$$ $$10x=9.999...$$ $$10x-9=0.999...$$ $$10x-9=x$$ $$10x-x=9$$ $$9x=9$$ $$x=1$$