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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 $x=1$.

Community
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kiss my armpit
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    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

4 Answers4

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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 $1$.

goblin GONE
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  • This made me understand why 0.999... admits a finite-decimal representation, while 0.333... does not! – PatrickT Mar 03 '18 at 07:04
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What I always find the most simple explanation is: $$ \frac{1}{3} = 0.333\ldots \quad \Longrightarrow \quad 1 = 3 \cdot \frac{1}{3} = 3 \cdot 0.333\ldots = 0.999\ldots $$

TMM
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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 $(1)$,$$1 = \underbrace{0.000\cdots}_0 + 0.999\cdots \\ 1 = 0.999\cdots$$

P.K.
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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 $0.999\ldots =\lim_{n\to\infty}q_n=1$.

Stefan Hansen
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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). –  Jan 05 '16 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 05 '16 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? –  Jan 05 '16 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 05 '16 at 15:53
  • By defining it to be equal to $1$. In fact, this was the philosophy for this post. –  Jan 05 '16 at 15:59
  • @user170039: Sure, you could also define to be equal to $3$. In my post I define to be the limit of the sequence $(q_n)$ or alternatively the number being $\leqslant 1$ and $\geqslant q_n$ for all $n$, and hence implicitly assuming that it's a real number. But so do you when you define to be equal to $1$. So what exactly is your point? – Stefan Hansen Jan 05 '16 at 16:03
  • Let us continue this discussion in chat. –  Jan 05 '16 at 16:05