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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 $x=1$.
On
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{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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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 $0.999\ldots =\lim_{n\to\infty}q_n=1$.
What I always find the most simple explanation is: $$ \frac{1}{3} = 0.333\ldots \quad \Longrightarrow \quad 1 = 3 \cdot \frac{1}{3} = 3 \cdot 0.333\ldots = 0.999\ldots $$