 Solution

LetThe tail satisfiesFor every fixed positive integer $d$, this is smaller than $q_N^{-d}$ for all sufficiently large $N$.

The binary expansion has $1$ digits at factorial positions and arbitrarily long blocks of zeros, but it is neither eventually zero nor eventually periodic. Since a rational number has an eventually periodic binary expansion, $\alpha$ is irrational. If it were an algebraic number of degree $d$, part (b), or equivalently the Liouville approximation theorem, would give a fixed positive multiple of $q^{-d}$ as a lower bound for all rational approximations. The fractions $p_N/q_N$ violate that bound. Thus $\alpha$ is a transcendental number.