Two questions:
Would the numeric string composed of concatenated primes be an irrational? Or w/o a true math proof, most surely?
Also, I am looking at the earliest prime terms seen as one progresses from each such concatenation. For example, from 7, we find 711,71113,7111317 are composite, then 711131719 is prime. From 5, we find 5711 as the prime. From 2 it is simply 23. I have limited computational capability, what is the soln starting from 3 eg 357111317...?
I have no clue. Thanks ...I have a Pari script if needed
For the first part: it is definitely irrational, as you could expect.
Suppose the contrary, and let $n$ be the length of the period. Since there is a finite number of digits before the period, let $q$ be the last prime we used that has some digits belonging to the antiperiod. Choose a big enough multiple $N=an$ such that $10^N>q$. Then at a certain point in the decimal expansion we would have $$\underbrace{\overline{p}\overline{p}\ldots \overline{p}}_{a\text{-times}}\ \underbrace{\overline{p}\overline{p}\ldots \overline{p}}_{a\text{-times}}$$ where $\overline{p}$ is a shifting of the digits of the period (if the period is $123$, then $\overline{p}$ could be $123$, $231$ or $312$), where the two terms are consecutive primes with $N$-digits. However this is clearly absurd, provided such primes exist.
Their existence it granted by Bertrand's postulate, which states that there is always a prime between $m$ and $2m$ for any $m>1$. Thus, there are always at least $3$ primes between $10^N$ and $10^{N+1}$ as $10^{N+1}>2^3\cdot 10^N$. This concludes the proof that the concatenation is irrational.
The part with Bertrand's postulate is somewhat of an overkill, but I can't think to an easier solution right now.
Edit: I didn't notice that Thomas Andrews had already posted a solution which is basically the same as mine. In case this is redundant, feel free to remove it.