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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 $x=1$.

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marked as duplicate by Did, Please don't touch, Asaf Karagila, David Mitra, dtldarek Jan 26 '13 at 14:15

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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 Answers 4

up vote 1 down vote accepted

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

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

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