# How can I explain $0.999\ldots=1$? [duplicate]

Possible Duplicate:
Does .99999… = 1?

I have to explain $0.999\ldots=1$ to people who don't know limit.

How can I explain $0.999\ldots=1$?

The common procedure is as follows

\begin{align} x&=0.999\ldots\\ 10x&=9.999\ldots \end{align}

$9x=9$ so $x=1$.

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## marked as duplicate by Did, kiss my armpit, Asaf Karagila, David Mitra, dtldarekJan 26 '13 at 14:15

what you show is a false 'proof' because you didn't write what 0.999... means and why 9.9999...-0.9999... = 9. To explain you need limit, otherwise don't explain. Giving someone a false proof is not necessarily better than leaving him with a doubt. – mez Jan 26 '13 at 15:29

What I always find the most simple explanation is: $$\frac{1}{3} = 0.333\ldots \quad \Longrightarrow \quad 1 = 3 \cdot \frac{1}{3} = 3 \cdot 0.333\ldots = 0.999\ldots$$

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It's a circular argument. – Stan Jun 6 '13 at 14:30

Between every two distinct real numbers, there exists a third real number distinct from the others. The contrapositive says that if no real numbers intermediate between $a$ and $b$, then $a$ equals $b$. So assume for a contradiction that a number intermediates between $0.9999...$ and $1$. For concreteness, lets say this number is $0.9981383...$ Well there is a first digit in this number that is not a 9. Thus $0.9981383...<0.99999...$. This contradicts the assumption that this number intermediates. Thus no number intermediates between $0.9999....$ and $1$.

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What do we understand when we see the number $0.999\ldots$? I understand the limit of sequence $(q_n)$ given by $$q_n=0.\underbrace{9\cdots9}_{n \text{ times}},\quad n\geq 1.$$ For a given $n$ the distance between $q_n$ and $1$ is $$|1-q_n|=0.\!\!\underbrace{0\cdots 0}_{n-1\text{ times}}\!\!1$$ which obvously goes to $0$ when $n$ tends to infinity. Hence $0.999\ldots =\lim_{n\to\infty}q_n=1$.

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This only implies that $(q_n)_{n\in \mathbb{N}}$ converges to $1$ whereas there seems to be no "meaning" to the symbol $0.999\ldots$ (unless, of course you define it to be "equal" to $1$, in which case there is nothing to prove). – user 170039 Jan 5 at 15:21
@user170039: I defined $0.999\ldots$ as the limit of the sequence $(q_n)$ as is written in the beginning of my post. Or you could say that $0.999\ldots$ is the number satisfying $0.999\ldots\leqslant 1$ and $0.999\ldots\geqslant q_n$ for all $n$ from which it follows that it's $1$. – Stefan Hansen Jan 5 at 15:42
My point is that you can only define $0.999\ldots$ to be the limit of $(q_n)$ if you know that it is a real number. What's the proof that it indeed is a real number? – user 170039 Jan 5 at 15:46
@user170039: If it's not given that it should be a real number, then you tell me how one should define or characterize it. That would be like saying 'Here is a quantity I'll call $x$, but I'm not willing to say anything about it. Now you tell me what it is'. – Stefan Hansen Jan 5 at 15:53
By defining it to be equal to $1$. In fact, this was the philosophy for this post. – user 170039 Jan 5 at 15:59

We know that if $a - b =0, ~~ a = b.$ So, we can do this:\begin{align}1 - 0.999\cdots & = 0.000\cdots \tag{1} \\ & = 0 \end{align}Rewriting $(1)$,$$1 = \underbrace{0.000\cdots}_0 + 0.999\cdots \\ 1 = 0.999\cdots$$

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