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.

1.0000.... and 0.99999...(infinitely many) are equivalent representation of the same real number — user, Feb 10 '18 at 11:02

score 2 · Answer 1 · answered Feb 10 '18 at 11:05

2

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

answered Feb 10 '18 at 11:05

idm

11,824

score 1 · Accepted Answer · answered Feb 10 '18 at 11:05

1

Classical proof:

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

answered Feb 10 '18 at 11:05

ə̷̶̸͇̘̜́̍͗̂̄︣͟

27,005

score 0 · Answer 3 · answered Feb 10 '18 at 11:04

0

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

answered Feb 10 '18 at 11:04

Olivier Oloa

120,989

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

3 Answers3