Repeat#2

Proving @Munem Sahariar 's solution.

$\begin{aligned} X & = 2.018\overline{19} \\ & = 2.018 + 0.000\overline{19} \\ & = 2.018 + 0.00019\left(1+10^{-2}+10^{-4} + \cdots \right) \\ & = 2.018 + 0.00019 \sum_{n=1}^\infty \left(\frac 1{100} \right)^n \\ & = 2.018 + 0.00019\left(\frac {100}{99}\right) \\ & = 2.018 + \frac {19}{99000} \\ & = \frac {2.018(99000) + 19}{99000} \\ & =\frac {2.018(100000-1000) + 19}{99000} \\ & = \color{#3D99F6} \frac {201819-2018}{99000} & \small \color{#3D99F6} \text{Munem Shariar's solution.} \\ & = \frac {199801}{99000} \end{aligned}$

Therefore, $2.018\overline{19}$ is rational .

$2.018\overline{19} = \frac{(201819 - 2018)}{99000} = \frac{199801}{99000}$

Hence it is rational.

$x = 2.018191919... \\ \begin{cases} 1000x = 2018.191919... \\ 100000x = 201819.191919... \end{cases}$

We can subtract the two equations, so we have:

$99000x = 199801 \\ x = \frac{199801}{99000}$

Hence $2.018 \overline{19}$ is rational .