Mills formula, $\theta^{3^n}$ is a prime for a certain $\theta$ and all natural $n$? I just watched this video done by numberphile, and the video claimed that there exist certain numbers $\theta$ such that the floor function of $\theta^{3^n}$ is a prime for all natural $n$ . $\theta$ is approximately $1.306$ . Doctor Grime (the guy presenting this) said that the best way to approximate $\theta$ right now is to is to find a Mills prime (such as 2) and solve for $\theta$ . If we do this, we get $\theta=\sqrt [3] 2$ , However, in another part of the video Dr. Grime said that it is unknown if $\theta$ is rational or irrational. How is this? It is pretty easy to prove that $\sqrt[3] 2$ is irrational: If $p$ and $q$ are co-prime, and $\frac pq=\sqrt[3] 2$ , then $\frac {p \cdot p \cdot p}{q\cdot q\cdot q\cdot}=2$ , however, this is a contradiction because we said $p$ and $q$ are comprime. So why would doctor Grime say that it is unknown if $\theta$ is irrational or not? Thanks.
P.S. Please try to keep the answers at a lower level (calculus 1).

Mills allows a certain degree of choice in the precise number chosen at each step. So, in truth, there are infinitely many numbers $\theta$ that work, and you do not know what the number is until you have chosen an infinite sequence of primes. That is, the theorem of Mills is cute, existence of the function is very nice, but is not a predictive tool. 
