Prove that $0.\overline{9}= 1$

Question

We all know that $\frac{1}{3} = 0.\overline{3}$, and that when we multiply it by $3$, we get $1$. But, $3\times \frac{1}{3} = 0.\overline{9}$.

Are there any ideas on how to prove that $0.\overline{9}= 1$? I have no ideas, I only wrote down that $0.\overline{9}+x= 1$ for $x\neq 0$, but I don't know how to proceed.

Answer 1

Classical proof:

Let $x=0.999...$ Then $$10x=9.999...\implies 10x-x=9.999...-0.999...$$ so $$9x=9\implies x=1$$

Answer 2

From $$ x=0.\overline{9} $$ one has $$ 10x=9+x,\qquad 9x=9, $$that is $$ x=0.\overline{9}=\frac{9}{9}=1. $$

Answer 3

$$0,999...=9\sum_{k=1}^\infty \frac{1}{10^n}=9\left(\frac{1}{1-\frac{1}{10}}-1\right)=1.$$