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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.

Jonathan Spirit
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  • Firstly, even if that denominator were a nonzero number, there would be no "asymptote," since the function is a linear function. And the slope would be $1/0.0\ldots 1$, which would be infinite, not $0.\overline{9}$. – Nishant Sep 06 '14 at 02:05
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    $0.000\ldots1$ is not a number that exists. So your friend's argument is based on a function that does not exist. One might as well argue that if you had a function that was a unicorn, the slope of the unicorn would be $0.999\ldots$ –  Sep 06 '14 at 02:18
  • It depends what '...' means. If it means 'the same forever' then 0.999... = 1. If it means 'we don't know what's next' then it doesn't equal one - use '..?' instead. –  Sep 06 '14 at 02:32
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    I think the biggest problem people have with things like .999...=1 is that they think of .999... as some intuitive object, that should behave in ways that are intuitive to them. What they don't understand is that it is in fact a completely formal object, that behaves in ways completely determined by the definitions. And, due to the formal definitions of "real numbers" being rather complicated to those unfamiliar with math, the way a number behaves may not at all be what someone expects if they don't know precisely what a "real number" is. – Carl Sep 06 '14 at 02:36
  • A good starting point to your next discussion is to ask them "what is the rigorous mathematical definition of an infinite decimal expansion?" If they can answer this then there is nothing left to argue about. – Winther Sep 06 '14 at 02:36
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    In this case, there is no point to argue. Whatever argument you give, independent of whether it is right or wrong, your friend won't listen because the argument is coming from you. You need some neutral respectable third party to repeat exactly the argument to convince them they are wrong. C'est la vie. – achille hui Sep 06 '14 at 02:37
  • Your friend is thinking of Brouwer's choice sequences, that's the alternative viewpoint. –  Sep 06 '14 at 02:38
  • Link to http://math.stackexchange.com/questions/11/is-0-999999999-ldots-1 – Caleb Stanford Jul 17 '16 at 01:43
  • Nothing —asymptotes or whatever— can disprove a true fact. – Mariano Suárez-Álvarez Oct 04 '16 at 02:08

9 Answers9

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$1/2 + 1/4 + 1/8 + \ldots$ 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.

  • LOL! I just commented about this on Hardy's answer! – Dair Sep 06 '14 at 02:26
  • Keep in mind that $1/2 + 1/4 + 1/8 ...$ is a limit. It is also a number, but the word limit is more informative. – Ryan Sep 06 '14 at 16:55
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    @Ryan: Maybe it is more informative in general, but giving more information can be a bad thing if it gets in the way of the information that's important, such as in the presence of the sort of confusion in the original question. –  Sep 06 '14 at 17:02
  • @Ryan: And maybe it's more accurate to say we can use the same notation to mean different things: sometimes we read at $1+2$ and the right thing to do is to interpret it as meaning the number $3$. Other times, we read $1+2$ and the right thing is to look at its parse tree. –  Sep 06 '14 at 17:03
  • @Hurkyl I was just emphasizing something which I thought lacked a lot of emphasis in your answer. I think your last sentence would make a much better first sentence in your answer - that is all. – Ryan Sep 06 '14 at 17:20
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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?

Michael Hardy
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David K
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$$x=0.999...$$ $$10x=9.999...$$ $$10x-9=0.999...$$ $$10x-9=x$$ $$10x-x=9$$ $$9x=9$$ $$x=1$$

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No, there are many valid proofs on wikipedia.

  • They reject each and every proof from there that I pull out, with wild arguments that are completely baseless. I'm stuck having to disprove every argument they throw my way. I know that all of those can be disproved, but I don't know how. – Jonathan Spirit Sep 06 '14 at 02:03
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    It's not necessary to argue against every wild claim. The burden is on them to refute the many proofs accepted by mathematicians, some of which are likely easily verifiable with something like Coq. –  Sep 06 '14 at 02:05
  • The problem is that they give me "counter-proofs" to the proofs I present, and then they hold fast to those counter-proofs even though they're wildly illogical. I'm thinking of just completely abandoning the argument. – Jonathan Spirit Sep 06 '14 at 02:11
  • If "they're wildly illogical", then point out how it is illogical. Pointing out any flaw in a proof invalidates the proof (it does not necessarily invalidate the result of the proof, but it does invalidate your claim that you have proved something). – Jared Sep 06 '14 at 02:15
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    get new friends, these ones sound absurd. – M A Pelto Sep 06 '14 at 02:18
  • Pointing out flaws is what I've been doing. Flaw after flaw, but they don't accept it and even come up with another terrible argument. – Jonathan Spirit Sep 06 '14 at 02:23
  • @JonathanSpirit They are completely wrong. If they bring up an argument and you have a counterargument, then this is all you can do. Now, because what you have stated is unequivocally fact, there should always be a problem with any proof that claims otherwise (and there always is). – Jared Sep 06 '14 at 02:24
  • @JonathanSpirit I think there are message boards where this thread may be more appropriate (you can present their arguments and people will likely comment shooting them down). – Jared Sep 06 '14 at 02:25
  • I've ended the argument. We're all irritating each other now, and I don't want that to happen. But thank you all for the reassurance that I'm right. – Jonathan Spirit Sep 06 '14 at 02:27
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"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.

Michael Hardy
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  • According to Spivak's Calculus: "This terminology is somewhat peculiar, because at best the symbol $\sum_{n=1}^{\infty} a_n$ denotes a number (so it can't "converge"), and it doesn't denote anything at all unless ${a_n}$ is summable. If I am interpreting Spivak correctly, you don't even need to worry about this whole "asymptotes" business, since when we put the = it is just defined that way. But, I could be very easily misunderstanding this. – Dair Sep 06 '14 at 02:25
  • @user667648: Spivak is right in his criticism, but the fact is that in practice the term "series" is ambiguous, and can mean either the expression one writes down, or the value that (in favourable cases) it denotes. It is hard to avoid sometimes talking about the concrete things we see before our eyes on paper (or imagine to see, because one cannot really write down a complete infinite series), rather than about their meaning (especially if there isn't any, as for divergent series). The same happens with "linear combinations" in linear algebra, which can be non-trivial even when they are zero. – Marc van Leeuwen Sep 06 '14 at 08:55
  • @MarcvanLeeuwen: Very true. Fortunately, I have yet to really meet any series that either interpretation really matters. Spivak also later mentions that it is a rather small technicality that is of little importance. – Dair Sep 06 '14 at 18:08
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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.

Jared
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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.

Michael Hardy
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Brendan Drew
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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 $1 = 0.99\ldots$

Michael Hardy
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Yes
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$0.9$ 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.

Irregular User
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Ratch Dux
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