Let $\alpha = 0.1011011101111\ldots$ be a given real number written in base 10

Question

Let $\alpha = 0.1011011101111\ldots$ be a given real number written in base $10$, that is, the n-th digit of $\alpha$ is $1$, unless n is of the form $\frac{k(k+1)}{2}-1$ in which case it is $0$. Choose all the correct statements from below.

1. $\alpha$ is a rational number
2. $\alpha$ is an irrational number
3. For every integer $q \geq 2$, there exists an integer $r \geq 1$ such that $\frac{r}{q} < \alpha < \frac{r+1}{q}$.
4. $\alpha$ has no periodic decimal expansion.

Answer 1

Hint : The length of the $1$-blocks is strictly increasing, so there cannot be a periodical expansion. This allows you to decide $1,2$ and $4$. To disprove $3$, just use $q=2$.

Answer 2

Number of $1$'s between $(\frac{k(k+1)}{2} -1)$th $0$ and $(\frac{(k+1)(k+2)}{2} -1)$th $0$ is $(\frac{k(k+1)}{2} - \frac{(k+1)(k+2)}{2}) = k+1$. i.e, number of ones between $0$'s are increasing and hence no periodic decimal expansion, so it is an irrational number. The other options are easy to counter.