0.999...=1/Proof 5

Theorem

$0.999 \ldots = 1$

Proof

\(\text {(1)}: \quad\)

\(\ds 0 . \underset n {\underbrace {999 \cdots 9} }\)

\(=\)

\(\ds 1 - 0.1^n\)

\(\ds 0.999 \cdots\)

\(=\)

\(\ds \eqclass {\sequence {0.9, \, 0.99, \, 0.999, \, \cdots} } {}\)

Definition of Real Numbers

\(\ds \)

\(=\)

\(\ds \eqclass {\sequence {1 - 0.1^1, \, 1 - 0.1^2, \, 1 - 0.1^3, \, \cdots} } {}\)

from $(1)$

\(\ds \)

\(=\)

\(\ds \eqclass {\sequence {1, \, 1, \, 1, \, \cdots} } {} - \eqclass {\sequence {0.1^1, \, 0.1^2, \, 0.1^3, \, \cdots} } {}\)

\(\ds \)

\(=\)

\(\ds 1 - 0\)

Sequence of Powers of Number less than One, Definition of Real Numbers

\(\ds \)

\(=\)

\(\ds 1\)

$\blacksquare$