Fun little problem

Assume that there is at least one two digit number $\overline{ab}$ such that $\overline{ba} = 2\overline{ab}$ .

It follows that $10b + a = 20a + 2b$ and therefore $a = \frac{8}{19}b$ .

One trivial solution is $b = 0$ also making $a = 0$ which we can't count as a solution as at least one of the two digits must be positive.

We will have infinitely many integer solutions for $a$ as long as $b$ is a multiple of $19$ . However we previously stated that $b$ is a digit and thus cannot equal $19$ or any of its multiples. We have reached a contradiction and therefore there is no two digit number $\overline{ab}$ such that $\overline{ba} = 2\overline{ab}$ and at least one of either $a$ or $b$ are positive.