As usual, excuse me for lack of latex. Assume rationality. Then there exist an integer d and number m < 10^d, x<10^d, such that all numbers of the form m + x*10^d*(10^(dy) - 1)/(10^d - 1) are in the sequence for all y. Such the problem statement, such a number is 0 modulo 2^((y+1)d) and by expanding we see that m(10^d - 1) - x*10^d must be zero modulo 2^((y+1)d). Since y can get big, this number must be 0 in general, but 10^d cannot divide m(10^d - 1) hence contradiction.

R8kt wrote: As usual, excuse me for lack of latex. Assume rationality. Then there exist an integer $d$ and number $m < 10^d, x<10^d$, such that all numbers of the form $m + x*10^d*\frac{10^{dy} - 1}{10^d - 1}$ are in the sequence for all $y$. Such the problem statement, such a number is $0$ modulo $2^{(y+1)d}$ and by expanding we see that $m(10^d - 1) - x*10^d$ must be zero modulo $2^{(y+1)d}$. Since $y$ can get big, this number must be 0 in general, but $10^d$ cannot divide $m(10^d - 1)$ hence contradiction.

Assume, for the sake of contradiction, that $r$ is rational. Then $r$ is a periodic decimal fraction. So $r=0. d_0d_1\ldots d_k(d_{k+1}d_{k+2}\ldots d_m)$. Call $a=d_k\ldots d_1d_0$ and $b=d_m\ldots d_{k+1}0\ldots 0$, where in the end we have $k+1$ zeros. Also denote $m-k=t$ - the length of the period. Now $e_n^2-e_n=e_n(e_n-1)\vdots 10^{n+1}$, but $e_n-1$ is odd, so $e_n \vdots 2^{n+1}$. We will consider only $n=k+1+At$, where $A$ is a positive integer. $e_n=a+b(1+10^t+\ldots+10^{(A-1)t})=a+\frac{b(10^{At}-1)}{10^t-1} \vdots 2^{n+1}$. Multiplying by $10^t-1$, we get $a(10^t-1)+b10^{At}-b \vdots 2^{n+1} \vdots 2^{At}$, so $a(10^t-1)-b \vdots 2^{At}$. But the LHS is a constant, while $A$ we can take arbitrarily large. Hence, $b=a(10^t-1)$. But $b$ has exactly $k+1$ zeros at the end, so $a$ too. But $a$ has length $k+1$, so $a=b=0$, which is a contradiction.