0.999...=1/Proof 5

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Theorem

$0.999 \ldots = 1$


Proof

\((1):\quad\) \(\displaystyle 0 . \underset n {\underbrace {999 \cdots 9} }\) \(=\) \(\displaystyle 1 - 0.1^n\)
\(\displaystyle 0.999 \cdots\) \(=\) \(\displaystyle \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}\) Definition of Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots} } {}\) from $(1)$
\(\displaystyle \) \(=\) \(\displaystyle \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {}\)
\(\displaystyle \) \(=\) \(\displaystyle 1 - 0\) Sequence of Powers of Number less than One, Definition of Real Numbers
\(\displaystyle \) \(=\) \(\displaystyle 1\)

$\blacksquare$